Convergence and Divergence of Series — Question 4

Question 4 – Solution

Step 1: Identify a benchmark.

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

Step 2: Compute the limit-comparison ratio.

$$\lim_{n\to\infty}\frac{(2n+1)/(n^3+1)}{1/n^2} =\lim_{n\to\infty}\frac{2n^3+n^2}{n^3+1}=2.$$ 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.

Step 3: Verify with a direct inequality.

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.