Consider the series

MATHEMATICS · Single correct · AP · Calculus BC · Infinite Sequences and Series · Infinite Sequences and Series

Problem

Consider the series ∑n=1∞nn3+1\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^{3}+1}. Which of the following comparisons is sufficient to conclude that this series converges? (a) Because for all n≥1,  nn3+1≥1n3   n\ge 1,\; \dfrac{n}{n^{3}+1}\ge \dfrac{1}{n^{3}}\; and ∑n=1∞1n3\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{3}} converges. (b) Because for all n≥1,  0≤nn3+1≤1n2   n\ge 1,\; 0\le\dfrac{n}{n^{3}+1}\le\dfrac{1}{n^{2}}\; and ∑n=1∞1n2\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{2}} converges. (c) Because for all n≥1,  0≤nn3+1≤1n   n\ge 1,\; 0\le\dfrac{n}{n^{3}+1}\le\dfrac{1}{n}\; and ∑n=1∞1n\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n} diverges. (d) Because for all n≥1,  0≤nn3+1≤1n   n\ge 1,\; 0\le\dfrac{n}{n^{3}+1}\le\dfrac{1}{\sqrt{n}}\; and ∑n=1∞1n\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}} diverges.

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