Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{n\left(n^{2}+1\right)} $$

Short answer

Therefore, the series \( \sum_{n=1}^{\infty} \frac{1}{n\left(n^{2}+1\right)}\) converges.

Step-by-step solution

Choose a Series to Compare To

Select a simpler series to compare with the original series. The simplest choice, considering the original series, is \( \sum_{n=1}^{\infty} \frac{1}{n^{3}}\). The chosen series is a p-series with p=3, which is known to converge because p > 1.

Apply the Limit Comparison Test

Compute the limit \( \lim_{{n\to\infty}}\frac{a_{n}}{b_{n}}\), where \(a_n= \frac{1}{n\left(n^{2}+1\right)} \) and \( b_n =\frac{1}{n^{3}}\). This simplifies to \( \lim_{{n\to\infty}}\frac{n^{2}}{n^{2}+1}\). As n approaches infinity, the +1 term becomes negligible, so this limit equals 1, which is a positive number.

Interpret the Result

Since the limit is a positive number (i.e., 1), according to the Limit Comparison Test, our original series will behave the same way as the comparison series. Hence, since our comparison series \( \sum_{n=1}^{\infty} \frac{1}{n^{3}}\) converges, then our original series \( \sum_{n=1}^{\infty} \frac{1}{n\left(n^{2}+1\right)}\) also converges.