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Chapter 7: Problem 56

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

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

Step by step solution

01

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.
02

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.
03

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.

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Most popular questions from this chapter

In your own words, define each of the following. (a) Sequence (b) Convergence of a sequence (c) Monotonic sequence (d) Bounded sequence

Suppose that \(\sum a_{n}\) and \(\sum b_{n}\) are series with positive terms. Prove that if \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\infty\) and \(\sum b_{n}\) diverges, \(\sum a_{n}\) also diverges.

Find all values of \(x\) for which the series converges. For these values of \(x,\) write the sum of the series as a function of \(x\). $$ \sum_{n=1}^{\infty}\left(\frac{x^{2}}{x^{2}+4}\right)^{n} $$

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