Use the integral test to determine whether the series converges.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 1 } { \sqrt { e ^ { 2 n } - 1 } }$$
A) converges
B) diverges

Question

Quizplus · Accepted Answer

We can use the integral test to determine if the given series converges or diverges. First, note that the terms of the series are positive for all $n\geq 1$. Let $f(x)= \frac{1}{\sqrt{e^{2x}-1}}$, then we have
$$\begin{aligned} \int_1^\infty{\frac{1}{\sqrt{e^{2x}-1}}dx} &\leq  \int_1^\infty{\frac{1}{\sqrt{e^{2x}}}dx} \ &=\int_1^\infty{\frac{1}{e^x}dx} \ &= \left. -e^{-x} ight|_1^{\infty} \ &=e^{-1}. \end{aligned}$$
Note that the inequality in the first line holds since the denominator is decreasing exponentially, and $e^{2x}-1\geq e^{2x}/2$ for all $x\geq 1$. Thus, we have
$$\sum_{n=1}^\infty \frac{1}{\sqrt{e^{2n}-1}} \leq \int_1^\infty{\frac{1}{\sqrt{e^{2x}-1}}dx} \leq e^{-1},$$
and since the integral on the right-hand side converges, the series on the left-hand side must also converge by the integral test. Therefore, the answer is $\boxed{	ext{(A)}}$.

Use the Integral Test to Determine Whether the Series Converges ∑n=1∞1e2n−1\sum _ { n = 1 } ^ { \infty } \frac { 1 } { \sqrt { e ^ { 2 n } - 1 } }n=1∑∞​e2n−1​1​

Use the Integral Test to Determine Whether the Series Converges $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { \sqrt { e ^ { 2 n } - 1 } }$