How to evaluate whether summation of $e^n/(ne^n+1)$ diverges or converges?

Question

How to test this summation for divergence or convergence? $$\sum_{n=0}^\infty \frac{e^n}{ne^n+1}$$

Edit: Here is my work, but I got it wrong. I tried using the comparison test.

\begin{align*} a_n &= \frac{e^n}{ne^n + 1} \\ &> \frac{1}{e^n \cdot e^n + e^{2n}} \\ &= \frac{1}{e^{2n} + e^{2n}} \\ &= \frac{1}{2e^{2n}} \\ &= \frac{1}{2}\left(\frac{1}{e}\right)^{2n} \\ b_n &= \frac{1}{e^n} &r= \frac{1}{e} < 1 \text{ converges due to geo-series} \end{align*} Since $b_n$ converges, $a_n$ also converges by CT.

Answer 1

An intuitive approach is that $e^n$ becomes huge, so divide the numerator and denominator by it. Then the $e^{-n}$ is tiny, so ignore it. Remember that convergence only depends on what happens for large $n$. Can you flesh this out?

Answer 2

Hint: factor $e^n$ out of the denominator, and compare it to $\frac{1}{n+1}$. What do you notice?

Edit: additionally, your reasoning is flawed because you showed that your first sequence (the one in the problem) is greater than $\frac{1}{2e^{2n}}$, and then you argue that because the smaller sequence converges, then so does the original, larger one. This makes no sense. You may as well started with $f(n)=n$, and concluded it converges by the same reasoning (which it certainly doesn't).

Answer 3

Since you're trying to understand what went wrong when you took this question on a test, I'll fill in the blanks in the other answers. Divide the numerator and denominator by $e^n$ to see that

$$\sum_{n=0}^\infty \frac{e^n}{ne^n+1}=\sum_{n=0}^\infty \frac{1}{n+e^{-n}} \gt \sum_{n=0}^\infty \frac{1}{n+1}.$$

The last sum diverges. Thus, you've now proved that this series is larger than a series that you know diverges, so the series must also diverge.