Infinite Series and Partial Sums

The partial-sum definition of a series is the entry point to nearly every quantitative science that relies on approximation. In numerical analysis, Taylor and power series are truncated partial sums used to approximate functions like $e^x$, $\sin x$, and $\log(1+x)$ to any desired precision, with the error controlled by exactly the kind of $\varepsilon$-$N$ estimates proved here. In probability and statistics, expectations of discrete random variables (e.g. for the Poisson or geometric distribution) are defined as sums of infinite series, and their convergence is what guarantees the expectation is well-defined. In physics and engineering, Fourier series decompose periodic signals into sums of sines and cosines, and questions of whether such a decomposition actually reconstructs the original signal are literal convergence-of-partial-sums questions. In cryptography and computer science, generating functions (formal power series) encode combinatorial counting problems, and convergence radius considerations determine which manipulations of these series are valid. In pure analysis, the entire theory of Geometric and Telescoping Series, the Divergence Test and Cauchy Criterion, and later tests (ratio, root, integral, comparison) are all elaborations built on exactly the two facts proved in this article: the necessity of $a_n \to 0$, and the equivalence between boundedness and convergence for nonnegative series.

Ask the tutor: How does the Cauchy criterion let us decide convergence of a series without ever knowing its limiting sum in advance?

Picture walking toward a wall, each step covering half the remaining distance: after step $N$ you have covered $1 - 2^{-N}$ of the total distance. You never reach the wall in finitely many steps, yet your position converges to it — and that limiting position is precisely what we mean when we declare $\sum 2^{-n} = 1$. The infinite sum is not a mystical completed addition; it is a name for the target of a sequence of ordinary, finite, entirely well-defined sums. This is why the definition of series convergence borrows nothing new from analysis: it is the $\varepsilon$-$N$ definition of limit, applied to one particular sequence — the running totals. The picture below shows the partial sums of $\sum 2^{-n}$ as points crowding toward $1$, and beside it the partial sums of the harmonic series $\sum 1/n$, which climb without bound even though each individual step $1/n \to 0$. The contrast is the entire content of the theorem that convergence requires $a_n \to 0$ but does not follow from it.

Proof technique used: direct application of the Monotone Convergence Theorem (Step 2) together with the elementary fact that convergent sequences are bounded (Step 3); both directions are proved without invoking any later series test, so there is no circularity.

Example 1 — a shifted, non-obvious geometric-type series computed directly from the definition. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)}$ converges, and if so find its sum, working directly from the definition of a series as a limit of partial sums (no formula for geometric series is used, since this is a telescoping series, foreshadowed below but computed here from first principles). Solution. Note the partial fraction decomposition

which can be checked by combining the right side over a common denominator: $\frac{(n+3)-(n+2)}{(n+2)(n+3)} = \frac{1}{(n+2)(n+3)}$, matching the left side exactly. Now compute the $N$-th partial sum by direct substitution and cancellation:

Every interior term cancels (each $-\frac{1}{k}$ from one bracket meets a $+\frac{1}{k}$ from the next), leaving only the first and last pieces:

We now find $\lim_{N\to\infty} s_N$ using the $\varepsilon$-$N$ definition of a limit directly. Fix $\varepsilon > 0$. We want $|s_N - \tfrac13| < \varepsilon$, i.e.

This holds whenever $N + 3 > 1/\varepsilon$, i.e. whenever $N > 1/\varepsilon - 3$. So choosing $N_0$ to be any integer exceeding $1/\varepsilon - 3$ (such $N_0$ exists by the Archimedean property of $\mathbb{R}$) verifies the definition: for all $N \ge N_0$, $|s_N - \frac13| < \varepsilon$. Hence $s_N \to \frac13$, and by definition the series converges with

As a sanity check: since all terms are positive, the theorem on nonnegative series confirms the sum should equal $\sup_N s_N$; indeed $s_N = \frac13 - \frac{1}{N+3}$ is increasing in $N$ and bounded above by $\frac13$, with supremum exactly $\frac13$, matching our computed limit.

Example 2 — using the Term Test to rule out convergence for a series that "looks" convergent. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n^2 + 3n + 1}{2n^2 + n + 5}$ converges. Solution. Rather than attempting to compute partial sums (there is no elementary closed form here), we apply the necessary condition proved above: if the series converges, its terms must tend to $0$. Compute the limit of the general term:

using the algebraic limit laws (dividing numerator and denominator by $n^2$, then using that $1/n \to 0$ and $1/n^2 \to 0$). Since the terms tend to $\tfrac12 \ne 0$, the contrapositive corollary (Divergence Test) applies directly: a series whose terms do not tend to $0$ cannot converge. Therefore

Notice what this example illustrates about proof strategy: we never needed to examine partial sums at all. The Term Test is often the fastest way to rule out convergence, though (as the harmonic series shows) it can never by itself establish convergence.