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Chapter 7: Problem 51

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

Short Answer

Expert verified
By using the Limit Comparison Test, it is found that the given series \(\sum_{n=1}^{\infty} \frac{n}{{n^{2}+1}}\) diverges.

Step by step solution

01

Identify the given series

The given series is \(\sum_{n=1}^{\infty} \frac{n}{{n^{2}+1}}\) . We need to determine whether this series converges or diverges.
02

Choose a suitable comparison series

We should choose a series that simplifies the given series but is known to converge or diverge. For this problem, a natural choice is the simpler series \(b_n = 1/n\). This is a p-series with \(p=1\), which is known to diverge.
03

Calculate the limit

Now, according to the Limit Comparison Test, find the limit of the ratio of \(a_n/b_n\) as \(n\) approaches infinity. \n Limit as \(n\) approaches infinity of \((n/(n^2+1))/(1/n)\) simplifies to \(\lim_ {n\rightarrow \infty} (n^2/(n^2+1))\). Using algebra we find that this limit is equal to \(1\), which is a finite number and greater than \(0\).
04

Conclude based on the Limit Comparison Test

The Limit Comparison Test states that if the limit is a finite number and greater than zero, then both series share the same nature. That is, if one series converges, the other does too. If one series diverges, the other does too. We've already figured out that our comparison series \(1/n\) diverges. Hence, our original series \(\sum_{n=1}^{\infty} \frac{n}{{n^{2}+1}}\) also diverges.

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Most popular questions from this chapter

State the definitions of convergent and divergent series.

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(n>1\), then \(n !=n(n-1) !\)

Write \(\sum_{k=1}^{\infty} \frac{6^{k}}{\left(3^{k+1}-2^{k+1}\right)\left(3^{k}-2^{k}\right)}\) as a rational number.

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(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0 . \overline{81} $$
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