Question

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

Short answer

The given series converges by the Limit Comparison Test with the comparison series \(\frac{1}{n^2}\).

Step-by-step solution

Identify the Comparison Series

Recognize that the denominator of the series \(\frac{1}{n \sqrt{n^{2}+1}}\) resembles a p-series structure. Select a comparison series or function that will simplify analysis. For this situation, select \(\frac{1}{n^2}\) as the comparison series because the terms of the given series and comparison series will simplify in limit calculations.

Apply the Limit Comparison Test

Apply the Limit Comparison Test by taking the limit as n approaches infinity of the ratio of the n-th term of the given series to the n-th term of the comparison series, which is \(\lim_{n\to\infty} \frac{\frac{1}{n \sqrt{n^{2}+1}}}{\frac{1}{n^2}}\). Simplify this limit to get \(\lim_{n\to\infty} \sqrt{n^2+1}\)

Calculate the Limit

Find the limit as n approaches infinity of \(\sqrt{n^2+1}\). As n goes to infinity, the limit will be infinity.

Interpret the Result

Since the limit is positive and finite (not zero), we can conclude that the given series has the same behavior as the comparison series. In other words, the given series will converge if and only if the comparison series converges. The comparison series \(\frac{1}{n^2}\) is a p-series with p = 2, which is greater than 1, therefore, the comparison series converges.