1
$\begingroup$

I'm trying to determine whether this series is convergent or divergent: $$ \sum_1^\infty \frac{e^{1/n}-1}{n} $$

I thought directly that $e^{1/n} \geq 1$ and therefore $\frac{e^{1/n}-1}{n} \geq \frac{1}{n}$, then using comparison test $\frac{e^{1/n}-1}{n}$ is divergent since $\frac{1}{n}$ is divergent, but apparently it converges! But I can't find what's wrong with my solution!

$\endgroup$
1
  • 3
    $\begingroup$ $e^{\frac{1}{n}} \geq 1$ implies that $e^{\frac{1}{n}} -1 \geq 0$, not $ \geq 1$ as you seem to be using. $\endgroup$ Oct 3, 2022 at 17:13

2 Answers 2

1
$\begingroup$

It is true that $e^{1/n}>1$. Then, it is true that $e^{1/n}-1>0$, but not true that $e^{1/n}-1>1$.

However, we do have for $n>1$, $e^{1/n}-1\le \frac1{1-1/n}-1=\frac1{n-1}$ (See THIS ANSWER). And clearly

$$\sum_{n=2}^\infty \frac{1}{n(n-1)}=\sum_{n=2}^\infty \frac1{n-1}-\frac1n=1$$ converges.

$\endgroup$
1
  • $\begingroup$ Would the downvoter care to comment? $\endgroup$
    – Mark Viola
    Oct 3, 2022 at 19:40
0
$\begingroup$

If $n$ is large $$a_n=\frac{e^{\frac 1n}-1}n=\frac 1{n^2}+\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$ Compare with $p$ series.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .