The Integral Test

Does the series ∑n=1∞1n\sum_{n=1}^\infty \frac{1}{n} converge? What about ∑n=1∞1n1.001\sum_{n=1}^\infty \frac{1}{n^{1.001}}? These two sums look almost identical. Yet one diverges to infinity while the other converges. No elementary comparison to a geometric series settles the matter. The Integral Test resolves this entire family of questions in one stroke. It does so by trading a discrete sum for a continuous improper integral. There, the machinery of antiderivatives becomes available. The idea matters because it converts an infinite discrete process into a different kind of question. Adding terms one at a time becomes a question about area under a curve. This is a question calculus already knows how to answer. This is one of the first deep bridges in analysis. It connects the discrete world of sequences and series. It connects that world to the continuous world of functions and integrals. It is the tool that finally pins down the convergence behavior of the enormously important p-series ∑1/np\sum 1/n^p. The big picture is geometric: if ff is a positive, decreasing function, the rectangles of height f(n)f(n) sit just above or just below the curve y=f(x)y = f(x). So the sum of rectangle areas and the area under the curve rise and fall together. Formalizing "rise and fall together" into a genuine theorem is the content of this article. Monotonicity and positivity serve as indispensable hypotheses. This article gives the precise statement and hypotheses of the Integral Test. It provides a complete proof by squeezing the partial sums between two integrals. It offers a full determination of p-series convergence as the flagship application. It includes a worked example estimating a tail sum. It also gives a counterexample showing why monotonicity cannot be dropped. It closes by placing the test in the broader landscape of convergence tests. This leads on to Absolute vs Conditional Convergence.

Key takeaways
  • The Integral Test compares ∑f(n)\sum f(n) to ∫1∞f(x) dx\int_1^\infty f(x)\,dx when ff is positive and decreasing.
  • Both objects converge or both diverge together — never one without the other.
  • It is proved by trapping partial sums between two integrals using monotonicity, then applying the Comparison Test logic to improper integrals.
  • Its flagship consequence: the p-series ∑1/np\sum 1/n^p converges iff p>1p > 1.
  • The same integral bounds give sharp estimates for how fast a series' tail shrinks, useful for numerical error control.
  • Key idea: monotone positive terms turn a sum into a Riemann-sum-like object trapped between two integrals of the same function.

Where this lives today

The Integral Test's core mechanism — bounding a discrete sum by a continuous integral of a monotone function — reappears throughout mathematics and its applications. In analytic number theory, the same squeeze inequality underlies Euler–Maclaurin summation and estimates for the growth of ∑n≤x1/n≈ln⁡x+γ\sum_{n\le x} 1/n \approx \ln x + \gamma (Euler's constant), and it is essential to understanding the analytic continuation and growth of the Riemann zeta function ζ(s)=∑n−s\zeta(s) = \sum n^{-s}, whose behavior near s=1s=1 is exactly governed by the pp-series threshold proved here. In probability and statistics, tail-sum/tail-integral comparisons of this type are used to bound the expectation of discrete random variables by comparing to continuous densities, and to control truncation error when a discrete distribution has a continuous analogue. In numerical analysis and scientific computing, the tail estimate derived from the test (as in Worked Example 2) is a standard device for certifying the accuracy of truncated series used to compute constants like ζ(3)\zeta(3) or special function values to guaranteed precision. In algorithm analysis (computer science), sums like ∑n=1N1/n\sum_{n=1}^N 1/n (harmonic numbers, appearing in the analysis of Quicksort and hashing) are routinely bounded above and below by ln⁡N\ln N using exactly this integral-comparison technique. Finally, within analysis itself, the Integral Test is a direct stepping stone toward finer convergence theory, including the study of Absolute vs Conditional Convergence for series whose terms are not all positive, and toward more refined tests (Cauchy condensation, Raabe's test) that handle series the Integral Test cannot reach because no elementary antiderivative exists.

theorem-quiz

Ask the tutor: How does the Integral Test's tail estimate compare in sharpness to the error bound given by the Alternating Series Test?

DefinitionThe Integral Test

Let NN be a positive integer, and let f:[N,∞)→Rf : [N, \infty) \to \mathbb{R} satisfy: 1. Positivity: f(x)≥0f(x) \geq 0 for all x≥Nx \geq N; 2. Monotonicity: ff is non-increasing on [N,∞)[N, \infty), i.e. x1≤x2  ⟹  f(x1)≥f(x2)x_1 \le x_2 \implies f(x_1) \ge f(x_2); 3. Continuity (or at least Riemann-integrability) of ff on every finite interval [N,M][N, M], so that ∫NMf(x) dx\int_N^M f(x)\,dx exists. Define an=f(n)a_n = f(n) for integers n≥Nn \geq N. The improper integral ∫N∞f(x) dx\int_N^\infty f(x)\,dx is defined as

∫N∞f(x) dx:=lim⁡M→∞∫NMf(x) dx,\int_N^\infty f(x)\,dx := \lim_{M \to \infty} \int_N^M f(x)\,dx,

and is said to converge if this limit exists (is finite), diverge otherwise. The Integral Test asserts that the series ∑n=N∞an\sum_{n=N}^\infty a_n and the improper integral ∫N∞f(x) dx\int_N^\infty f(x)\,dx converge or diverge together.

Intuition

Picture the graph of a positive, decreasing function ff on [1,∞)[1,\infty). Above each integer nn, erect a rectangle of width 11 and height f(n)f(n). Because ff is decreasing, the rectangle built on [n,n+1][n, n+1] with height f(n)f(n) — the left endpoint value — sits entirely above the curve on that interval, while the rectangle with height f(n+1)f(n+1) sits entirely below the curve. Stacking these observations across all intervals produces the chain of inequalities that drives the whole theorem: the partial sums of f(n)f(n) overestimate the running integral, and the partial sums of f(n+1)f(n+1) underestimate it. Since both sums differ from the integral only by bounded, controllable amounts, they must diverge or converge in lockstep with it. This picture immediately explains the p-series threshold at p=1p=1: ∫1∞x−p dx\int_1^\infty x^{-p}\,dx is an elementary computation, and it converges exactly when p>1p>1 because the antiderivative x1−p/(1−p)x^{1-p}/(1-p) tends to a finite limit only when the exponent 1−p1-p is negative.

:::svg[Rectangles over and under y=1/xy = 1/x on [1,5][1,5], illustrating the two-sided squeeze] x y 1 2 3 4 5 y = 1/x orange: f(n) rectangles (over) green: f(n+1) rectangles (under)

TheoremThe Integral Test

Let N∈Z+N \in \mathbb{Z}^+ and let f:[N,∞)→Rf:[N,\infty) \to \mathbb{R} be positive, non-increasing, and Riemann-integrable on every finite subinterval [N,M][N,M]. Set an=f(n)a_n = f(n) for n≥Nn \geq N. Then:

∑n=N∞an converges  ⟺  ∫N∞f(x) dx converges.\sum_{n=N}^\infty a_n \ \text{converges} \iff \int_N^\infty f(x)\,dx \ \text{converges}.

Moreover, when both converge, for every integer K≥NK \geq N the tail satisfies

∫K+1∞f(x) dx ≤ ∑n=K+1∞an ≤ ∫K∞f(x) dx.\int_{K+1}^\infty f(x)\,dx \ \le \ \sum_{n=K+1}^\infty a_n \ \le \ \int_K^\infty f(x)\,dx.
Proof

We use the technique of squeezing partial sums between two Riemann sums built from monotonicity, then invoke the monotone convergence of bounded increasing sequences.

Step 1: The key inequality. Fix any integer n≥Nn \geq N. Since ff is non-increasing on [n,n+1][n, n+1], for every x∈[n,n+1]x \in [n, n+1] we have

f(n+1)≤f(x)≤f(n).f(n+1) \leq f(x) \leq f(n).

Integrating this inequality over x∈[n,n+1]x \in [n, n+1] (a valid operation since integration preserves order of integrable functions on an interval of length 1) gives

f(n+1)=∫nn+1f(n+1) dx ≤ ∫nn+1f(x) dx ≤ ∫nn+1f(n) dx=f(n).f(n+1) = \int_n^{n+1} f(n+1)\,dx \ \le \ \int_n^{n+1} f(x)\,dx \ \le \ \int_n^{n+1} f(n)\,dx = f(n).

That is,

an+1≤∫nn+1f(x) dx≤an.(∗)a_{n+1} \le \int_n^{n+1} f(x)\,dx \le a_n. \tag{$\ast$}

Step 2: Sum (∗)(\ast) from n=Nn = N to n=M−1n = M-1 for an arbitrary integer M>NM > N. Summing the left inequality:

∑n=NM−1an+1≤∑n=NM−1∫nn+1f(x) dx=∫NMf(x) dx,\sum_{n=N}^{M-1} a_{n+1} \le \sum_{n=N}^{M-1} \int_n^{n+1} f(x)\,dx = \int_N^M f(x)\,dx,

where the equality uses additivity of the integral over adjacent intervals. The left side telescopes to ∑k=N+1Mak\sum_{k=N+1}^{M} a_k. So:

∑k=N+1Mak ≤ ∫NMf(x) dx.(A)\sum_{k=N+1}^{M} a_k \ \le \ \int_N^M f(x)\,dx. \tag{A}

Summing the right inequality of (∗)(\ast) similarly:

∫NMf(x) dx ≤ ∑n=NM−1an.(B)\int_N^M f(x)\,dx \ \le \ \sum_{n=N}^{M-1} a_n. \tag{B}

Combining (A) and (B):

∑k=N+1Mak ≤ ∫NMf(x) dx ≤ ∑n=NM−1an.(C)\sum_{k=N+1}^{M} a_k \ \le \ \int_N^M f(x)\,dx \ \le \ \sum_{n=N}^{M-1} a_n. \tag{C}

Step 3: Both sequences (partial sums, partial integrals) are non-decreasing in MM. Because f≥0f \geq 0, each an≥0a_n \geq 0, so the partial sums SM:=∑n=NManS_M := \sum_{n=N}^{M} a_n form a non-decreasing sequence in MM. Likewise IM:=∫NMf(x) dxI_M := \int_N^M f(x)\,dx is non-decreasing in MM, since IM+1−IM=∫MM+1f(x) dx≥0I_{M+1} - I_M = \int_M^{M+1} f(x)\,dx \geq 0 (the integral of a non-negative function is non-negative).

Step 4: (⇐\Leftarrow) Suppose ∫N∞f(x) dx\int_N^\infty f(x)\,dx converges, i.e. IM→LI_M \to L for some finite LL as M→∞M \to \infty (equivalently, since IMI_M is non-decreasing, IM≤LI_M \le L for all MM). By inequality (A),

SM−aN=∑k=N+1Mak≤IM≤LS_M - a_N = \sum_{k=N+1}^M a_k \le I_M \le L

for every M>NM > N, so SM≤L+aNS_M \le L + a_N for all MM. Thus (SM)(S_M) is a non-decreasing sequence bounded above, hence convergent by the Monotone Convergence Theorem for sequences (a non-decreasing sequence bounded above converges to its supremum). Since SM→∑n=N∞anS_M \to \sum_{n=N}^\infty a_n by definition of series convergence, the series converges.

Step 5: (⇒\Rightarrow) Suppose ∑n=N∞an\sum_{n=N}^\infty a_n converges, i.e. SM→SS_M \to S for some finite SS, so in particular (SM)(S_M) is bounded above by SS (as it is non-decreasing with limit SS, every term is ≤S\le S). By inequality (B),

IM≤SM−1≤SI_M \le S_{M-1} \le S

for all M>NM > N. Thus (IM)(I_M) is non-decreasing and bounded above, hence converges by the same Monotone Convergence Theorem. Since IM→∫N∞f(x) dxI_M \to \int_N^\infty f(x)\,dx by definition, the improper integral converges. This establishes the equivalence. (Divergence in one direction is the contrapositive: if the integral diverges, i.e. IM→∞I_M \to \infty, then Step 4's chain shows SM≥IM−aN→∞S_M \geq I_M - a_N \to \infty hmm — more directly: if IM→∞I_M\to\infty then by (A), since IM≤SMI_M \le S_M actually we use (B) rearranged: IM≤SM−1I_M \le S_{M-1}, so SM−1≥IM→∞S_{M-1} \geq I_M \to \infty, forcing SM→∞S_M \to \infty too, i.e. the series diverges. Symmetrically if SM→∞S_M \to \infty, (A) gives IM≥SM−aN→∞I_M \geq S_M - a_N \to \infty.)

Step 6: The tail estimate. Fix K≥NK \geq N. Apply inequality (C) with the roles of the lower limit shifted to start at KK: for n≥Kn \geq K, summing (∗)(\ast) from n=Kn=K to M−1M-1 and letting M→∞M \to \infty (valid termwise since all sums/integrals involved converge, by the equivalence just proved) gives

∑n=K+1∞an ≤ ∫K∞f(x) dx,\sum_{n=K+1}^{\infty} a_n \ \le \ \int_K^{\infty} f(x)\,dx,

directly from the limiting form of (B) started at KK, and

∫K+1∞f(x) dx ≤ ∑n=K+1∞an,\int_{K+1}^{\infty} f(x)\,dx \ \le \ \sum_{n=K+1}^{\infty} a_n,

from the limiting form of (A) started at KK. Combining gives exactly the stated tail bound. ■\blacksquare

∎
CorollaryConvergence of the p-series

For p∈Rp \in \mathbb{R}, the series ∑n=1∞1np\displaystyle\sum_{n=1}^\infty \frac{1}{n^p} converges if p>1p > 1 and diverges if p≤1p \le 1.

Proof

Case p≤0p \le 0: then 1/np=n−p≥11/n^p = n^{-p} \geq 1 for all n≥1n \geq 1 (since −p≥0-p \geq 0), so the terms do not tend to 00; by the Term Test (divergence when terms fail to vanish, itself an elementary consequence of the Cauchy criterion for series), the series diverges. This case needs no integral.

Case p>0p > 0: the function f(x)=x−pf(x) = x^{-p} is positive, continuous, and strictly decreasing on [1,∞)[1,\infty) (since p>0p>0), so the hypotheses of the Integral Test hold with N=1N=1. Compute, for M>1M > 1: If p≠1p \neq 1:

∫1Mx−p dx=[x1−p1−p]1M=M1−p−11−p.\int_1^M x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_1^M = \frac{M^{1-p} - 1}{1-p}.

As M→∞M \to \infty: if p>1p > 1, then 1−p<01-p < 0, so M1−p→0M^{1-p} \to 0, giving ∫1∞x−p dx=−11−p=1p−1\int_1^\infty x^{-p}\,dx = \frac{-1}{1-p} = \frac{1}{p-1}, a finite value — the integral converges. If 0<p<10 < p < 1, then 1−p>01-p>0, so M1−p→∞M^{1-p}\to\infty, and the integral diverges. If p=1p = 1:

∫1Mx−1 dx=ln⁡M→∞(M→∞),\int_1^M x^{-1}\,dx = \ln M \to \infty \quad (M\to\infty),

so the integral diverges. By the Integral Test, ∑1/np\sum 1/n^p converges exactly when ∫1∞x−p dx\int_1^\infty x^{-p}\,dx converges, i.e. exactly when p>1p>1. Combined with the p≤0p\le 0 case, and noting 0<p≤10<p\le1 diverges by the computation above, we conclude convergence iff p>1p>1. ■\blacksquare

∎

Worked examples

Example 1 — A non-trivial p-series-type comparison: ∑n=2∞1n(ln⁡n)2\sum_{n=2}^\infty \dfrac{1}{n(\ln n)^2}. Let f(x)=1x(ln⁡x)2f(x) = \dfrac{1}{x(\ln x)^2} for x≥2x \geq 2. Check the hypotheses:

  • Positivity: for x≥2x \geq 2, x>0x>0 and (ln⁡x)2>0(\ln x)^2 > 0 (since ln⁡2≠0\ln 2 \ne 0), so f(x)>0f(x) > 0.
  • Monotonicity: both xx and (ln⁡x)2(\ln x)^2 are increasing for x≥2x \geq 2 (as ln⁡x≥ln⁡2>0\ln x \ge \ln 2 >0, so (ln⁡x)2(\ln x)^2 is increasing there too), hence their product is increasing, so f(x)=1/[x(ln⁡x)2]f(x) = 1/[x(\ln x)^2] is decreasing.
  • Continuity: ff is a composition/product of continuous functions on [2,∞)[2,\infty), hence continuous, hence Riemann-integrable on every [2,M][2,M].

Compute the improper integral using the substitution u=ln⁡xu = \ln x, du=dx/xdu = dx/x:

∫2Mdxx(ln⁡x)2=∫ln⁡2ln⁡Mduu2=[−1u]ln⁡2ln⁡M=1ln⁡2−1ln⁡M.\int_2^M \frac{dx}{x(\ln x)^2} = \int_{\ln 2}^{\ln M} \frac{du}{u^2} = \left[-\frac{1}{u}\right]_{\ln 2}^{\ln M} = \frac{1}{\ln 2} - \frac{1}{\ln M}.

As M→∞M \to \infty, ln⁡M→∞\ln M \to \infty so 1/ln⁡M→01/\ln M \to 0, giving

∫2∞dxx(ln⁡x)2=1ln⁡2≈10.6931≈1.4427,\int_2^\infty \frac{dx}{x(\ln x)^2} = \frac{1}{\ln 2} \approx \frac{1}{0.6931} \approx 1.4427,

a finite value. By the Integral Test, ∑n=2∞1n(ln⁡n)2\sum_{n=2}^\infty \dfrac{1}{n(\ln n)^2} converges. This example is instructive precisely because no direct comparison to a p-series works cleanly (1/[n(ln⁡n)2]1/[n(\ln n)^2] is asymptotically smaller than 1/n1/n but the ratio to 1/n1+ϵ1/n^{1+\epsilon} has no clean limit for small ϵ\epsilon), while the antiderivative computation is immediate. It also illustrates the general fact that ∑1/[n(ln⁡n)q]\sum 1/[n(\ln n)^q] converges iff q>1q > 1, by the identical substitution — a companion family to the p-series often called Bertrand series.

Example 2 — Using the tail estimate to bound an error in a numerical approximation. Consider ∑n=1∞1n3\sum_{n=1}^\infty \dfrac{1}{n^3} (a convergent p=3p=3 series). Suppose we approximate the sum S=∑n=1∞n−3S = \sum_{n=1}^\infty n^{-3} by the partial sum

S10=∑n=1101n3,S_{10} = \sum_{n=1}^{10} \frac{1}{n^3},

and we want a rigorous bound on the error S−S10=∑n=11∞n−3S - S_{10} = \sum_{n=11}^\infty n^{-3}. Apply the tail estimate from the Integral Test with f(x)=x−3f(x) = x^{-3} (positive, decreasing on [1,∞)[1,\infty)) and K=10K = 10:

∫11∞x−3 dx ≤ ∑n=11∞1n3 ≤ ∫10∞x−3 dx.\int_{11}^\infty x^{-3}\,dx \ \le \ \sum_{n=11}^\infty \frac{1}{n^3} \ \le \ \int_{10}^\infty x^{-3}\,dx.

Compute both integrals using ∫x−3 dx=−12x−2\int x^{-3}\,dx = -\tfrac{1}{2}x^{-2}:

∫11∞x−3 dx=12⋅112=1242≈0.0041322,\int_{11}^\infty x^{-3}\,dx = \frac{1}{2 \cdot 11^2} = \frac{1}{242} \approx 0.0041322,
∫10∞x−3 dx=12⋅102=1200=0.005.\int_{10}^\infty x^{-3}\,dx = \frac{1}{2 \cdot 10^2} = \frac{1}{200} = 0.005.

Hence

0.0041322 ≲ ∑n=11∞1n3 ≲ 0.005,0.0041322 \ \lesssim \ \sum_{n=11}^\infty \frac{1}{n^3} \ \lesssim \ 0.005,

so the true sum SS satisfies S10+0.0041322≤S≤S10+0.005S_{10} + 0.0041322 \le S \le S_{10} + 0.005. Computing S10=1+18+127+164+1125+1216+1343+1512+1729+11000≈1.19753S_{10} = 1 + \tfrac18 + \tfrac{1}{27}+\tfrac{1}{64}+\tfrac{1}{125}+\tfrac{1}{216}+\tfrac{1}{343}+\tfrac{1}{512}+\tfrac{1}{729}+\tfrac{1}{1000} \approx 1.19753, we obtain the rigorous bracket

1.20166 ≲ S=∑n=1∞1n3 ≲ 1.20253,1.20166 \ \lesssim \ S = \sum_{n=1}^\infty \frac{1}{n^3} \ \lesssim \ 1.20253,

consistent with the known value ζ(3)≈1.2020569\zeta(3) \approx 1.2020569. This demonstrates the practical power of the tail-estimate corollary: it converts the qualitative convergence statement into a quantitative, computable error bound, entirely from elementary antiderivatives, with no need to sum extra terms.

CounterexampleWhy monotonicity cannot be dropped

Consider f:[1,∞)→Rf: [1,\infty) \to \mathbb{R} defined by

f(x)={1if x∈[n,n+n−2] for some integer n≥1,0otherwise.f(x) = \begin{cases} 1 & \text{if } x \in [n, n + n^{-2}] \text{ for some integer } n \ge 1, \\ 0 & \text{otherwise.} \end{cases}

This ff is positive (well, non-negative — take f≥0f\ge 0, positivity in the weak sense used implicitly) and Riemann-integrable on every [1,M][1,M] (it is a finite union of intervals where it's 11, else 00, for xx in any bounded range), but it is not monotone: it oscillates between spikes of height 11 and stretches of height 00 infinitely often. Define an=f(n)=1a_n = f(n) = 1 for every integer n≥1n \geq 1 (since each integer nn sits at the left endpoint of a spike [n,n+n−2][n, n+n^{-2}]). Then

∑n=1∞an=∑n=1∞1=∞diverges.\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty 1 = \infty \quad \text{diverges.}

But the integral only picks up the narrow spikes:

∫1∞f(x) dx=∑n=1∞n−2=π26<∞converges.\int_1^\infty f(x)\,dx = \sum_{n=1}^\infty n^{-2} = \frac{\pi^2}{6} < \infty \quad \text{converges.}

So the series diverges while the integral converges — the conclusion of the Integral Test fails entirely once monotonicity is dropped. The proof's Step 1 inequality (∗)(\ast), which pins f(x)f(x) between f(n)f(n) and f(n+1)f(n+1) on [n,n+1][n,n+1], is simply false for this oscillating ff, since ff can be 11 at the sample point nn but 00 almost everywhere else on [n,n+1][n,n+1]; the entire squeeze argument collapses. This shows monotonicity is not a technical convenience but load-bearing: it is what lets a single sample value f(n)f(n) control the integral over the whole unit interval [n,n+1][n,n+1].

Scope, edge cases & common traps

SituationWhat to checkCommon pitfall
Series starts at n=1n=1 but ff only decreasing from n=5n=5 onwardApply the test on [5,∞)[5,\infty); the first four terms are a finite sum, irrelevant to convergenceTrying to apply the test on all of [1,∞)[1,\infty) and failing because ff isn't monotone there
ff decreasing but not continuous (e.g. a step function)Riemann-integrability on finite intervals suffices; continuity is not requiredAssuming the test needs continuity — it only needs integrability
Series with negative or alternating termsIntegral Test does not apply — positivity is essentialMisapplying the test to ∑(−1)n/n\sum (-1)^n/n; the correct tool is the Alternating Series Test, feeding into Absolute vs Conditional Convergence
Non-increasing but eventually constant ff (e.g. f(x)=0f(x) = 0 for xx large)Still valid; both sum and integral trivially convergeNone — this case is often mistakenly thought to need special handling
Using tail estimate to bound errorMust use the same KK consistently in both integral boundsOff-by-one errors: using ∫K∞\int_K^\infty paired with ∑n=K∞\sum_{n=K}^\infty instead of ∑n=K+1∞\sum_{n=K+1}^\infty

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Early 19th-century French analysis, École Polytechnique

Cauchy is the figure most closely associated with giving rigorous, quantifier-precise foundations to the convergence of series and integrals, replacing 18th-century formal manipulation with ε\varepsilon-based definitions. The comparison between a monotone series and its associated integral appears in his systematic development of convergence criteria in his Cours d'analyse (1821) and later lecture notes, part of his broader program of rebuilding calculus on solid logical footing after the looser methods of Euler and the Bernoullis.

History
- 1650s: Pietro Mengoli and later Jacob Bernoulli study ∑1/n\sum 1/n (the harmonic series) and prove its divergence by grouping terms, without any integral comparison.
- 1730s: Leonhard Euler computes ∑1/n2=π2/6\sum 1/n^2 = \pi^2/6 and studies ∑1/np\sum 1/n^p for various pp, using formal, non-rigorous analytic techniques including early forms of the Euler–Maclaurin summation formula.
- 1821: Augustin-Louis Cauchy publishes Cours d'analyse, formalizing limits and convergence with modern rigor and stating comparison principles between series and integrals.
- 1830s: Cauchy and Peter Gustav Lejeune Dirichlet refine convergence criteria; the integral comparison becomes a standard tool alongside the ratio and root tests.
- Late 19th century: The test is codified in textbooks in essentially its modern form, alongside the emerging Riemann and later Lebesgue theories of integration that make "improper integral" fully precise.
Does the Integral Test tell us the exact value of the sum?

No. It only tells you whether the series converges or diverges, by comparing to whether the improper integral converges or diverges — the numerical value of the integral is generally not equal to the sum. For example ∫1∞x−2 dx=1\int_1^\infty x^{-2}\,dx = 1 exactly, but ∑n=1∞n−2=π2/6≈1.645\sum_{n=1}^\infty n^{-2} = \pi^2/6 \approx 1.645, a different number. What the test's proof does give you, via the tail-estimate corollary, is a rigorous two-sided bound on the tail of the series in terms of integrals, which can be used to approximate the sum to any desired accuracy — but the integral itself is not the sum.

Why does the function need to be defined on a continuous interval, not just at integers?

The entire proof rests on integrating the inequality f(n+1)≤f(x)≤f(n)f(n+1) \le f(x) \le f(n) over the real interval [n,n+1][n, n+1], which requires ff to be a genuine function of a real variable, integrable on that interval. Without this continuous extension, there is no integral to compare the series to. The choice of which continuous ff to use is not unique — any positive, non-increasing continuous extension of the sequence ana_n works equally well, since the test only depends on the values f(n)=anf(n)=a_n and the monotonicity/positivity structure, not on the specific formula for ff between integers.

Can the Integral Test be used when ff is increasing instead of decreasing?

Not directly in the stated form, but a symmetric version exists: for a positive, non-decreasing ff, the inequalities in Step 1 simply reverse (f(n)≤f(x)≤f(n+1)f(n) \le f(x) \le f(n+1)), and the same telescoping argument shows ∑f(n)\sum f(n) and ∫f\int f still converge or diverge together — but such a series almost never converges anyway, since a non-decreasing positive sequence either tends to a nonzero limit (violating the Term Test) or grows further, so this case is rarely useful in practice. The genuinely useful, standard form is for decreasing ff, which is why textbooks state it that way.

How does the Integral Test relate to the Comparison Test?

They share the same underlying philosophy — bound one series-like object by another whose behavior you already understand — but they operate differently. The Comparison Test compares two series termwise using an inequality 0≤an≤bn0 \le a_n \le b_n. The Integral Test instead compares a series to a continuous integral built from the same formula, using monotonicity rather than an external comparison series. In practice, the Integral Test is often used first to establish convergence facts about p-series, which the Comparison Test (or its limit form) then uses as reference series for comparing more complicated series that are not directly integrable in closed form.

What if the antiderivative of ff has no elementary closed form?

The Integral Test only requires determining whether lim⁡M→∞∫NMf(x) dx\lim_{M\to\infty}\int_N^M f(x)\,dx is finite or infinite — you do not need a closed-form antiderivative, only enough analytic control (asymptotic estimates, further integral comparisons, or known special-function behavior) to determine convergence of the improper integral. In many textbook applications the antiderivative is elementary (as with x−px^{-p}), but the test's logical content survives even when it is not; one simply falls back on other techniques to analyze the improper integral itself.

Is it true that if ∑an\sum a_n converges then an→0a_n \to 0 "fast enough" that the Integral Test always applies?

No — this reverses the logic. The Integral Test is a sufficient tool for series that happen to arise from a positive, decreasing function; it is not a universal method for all convergent positive series. Many convergent series (e.g. one with an=1/n2a_n = 1/n^2 for even nn and an=1/2na_n = 1/2^n for odd nn) have terms that are not monotone in nn, so no single decreasing ff with f(n)=anf(n)=a_n need behave simply, and the test's hypotheses may fail to hold in a useful form even though the series converges. Convergence of an→0a_n\to 0 is necessary but never sufficient for convergence of ∑an\sum a_n, and it says nothing about monotonicity.

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