Determine whether or not $\sum_{k=1}^{\infty} \frac{1}{k- \mathrm{e}^{-k}}$ converges.

Question

I have the following so far

Let $a_k = \frac{1}{k- e^{-k}}$.

Now, $\lim_{k \to \infty}a_k=0 \implies$ $\sum a_k$ can either converge or diverege. We must thus do further tests to determine whether or not our infinite series diverges.

Let $f:[1,\infty)\to\mathbb{R}$ be defined by $f(t)=\frac{1}{t-e^{-t}}$.

Let $g(t)=t-e^{-t}$ denote our denominator of $f(t)$. Now $g'(t)=1+e^{-t} \geq0 \ \forall t\in \mathbb{R}$. Thus we have that $g(t)$ is increasing and $f(t)$ is decreasing. Thus the criteria to use the Integral Test is satisfied.

Integral Test:

\begin{align*}\int \limits_{1}^{\infty} \frac{1}{k-e^{-k}}dk &= \lim_{a\to \infty} \int \limits_{1}^{a}\frac{1}{k-e^{-k}}dk\end{align*} Which does not converge (I used Mathematica to show this, as I am unable to integrate this by hand).

Hence we have that $\sum a_k$ diverges.

Is this method correct, or is there another way of doing that which would not require me to use Mathematica for the integral, as I will not be able to do so in a test environment.

Answer 1

Note that $$ \frac{1}{k}<\frac{1}{k-e^{-k}}, $$ and as $\sum_{k=1}^\infty\frac{1}{k}$ diverges, so does $\sum_{k=1}^\infty\frac{1}{k-e^{-k}}$.

Answer 2

You can simply use comparison test. You know that for large $k$ you have:

$$e^{-k}<\frac{k}{2}$$

(for example by computing limit $\lim_{x \to \infty}e^{-x}/x$)

So for large $k$

$$\frac{1}{k-e^{-k}}>\frac{1}{k-\frac{k}{2}}=\frac{2}{k}$$

But series $\sum_{n=1}^{\infty}\frac{2}{n}$ diverges.

But your method is also correct.