Use the Limit Comparison Test to determine the convergence or divergence of the series .
A)The series diverges.
B)The series converges.

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series  . ​
A)The series diverges. 
B)The series converges.

Quizplus · Accepted Answer

We use the limit comparison test with the series $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}$.

Note that $n\sqrt{n+1} \geq n\sqrt{n} = n^{3/2}$ for all $n$, so $\frac{1}{n\sqrt{n+1}} \leq \frac{1}{n^{3/2}}$.

Using the limit comparison test with $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$, we have

$$\lim_{n 	o \infty} \frac{\frac{1}{n\sqrt{n+1}}}{\frac{1}{n^{3/2}}} = \lim_{n 	o \infty} \frac{n^{3/2}}{n\sqrt{n+1}} = \lim_{n 	o \infty} \frac{1}{\sqrt{1 + \frac{1}{n}}} = 1$$

Since $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges (by the p-series test with $p=\frac{3}{2}$), we conclude that $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}$ converges by the limit comparison test.

Therefore, the answer is choice B.

Use the Limit Comparison Test to Determine the Convergence or Divergence