Convergence and Divergence of Series — Question 4

*Question 4 Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{2n+1}{n^3+1}$ converges or diverges.

1. Identify the reciprocal-power benchmark suggested by the leading terms.

2. Compute the limit-comparison ratio with that benchmark.

3. State the conclusion and give a direct upper comparison valid for $n\ge1$.

*Question 4 – Solution

The terms are positive. Comparing the highest powers gives $(2n)/(n^3)=2/n^2$, so the natural benchmark is the convergent $p$-series $b_n=1/n^2$.

$$\lim_{n\to\infty}\frac{(2n+1)/(n^3+1)}{1/n^2} =\lim_{n\to\infty}\frac{2n^3+n^2}{n^3+1}=2.$$ The ratio limit is $2$, which is finite and strictly positive. These are exactly the hypotheses of the Limit Comparison Test. Since $\sum1/n^2$ converges ($p=2>1$), the given series converges.

For $n\ge1$, $2n+1\le3n$ and $n^3+1\ge n^3$, so $$0<\frac{2n+1}{n^3+1}\le\frac3{n^2}.$$ Because $\sum3/n^2$ converges, the Direct Comparison Test independently confirms the conclusion. The lower-order terms affect numerical values but not tail behavior.