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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.

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    $\begingroup$ What have you already tried? Why is this problem of interest to you? $\endgroup$ Mar 17, 2019 at 3:27
  • $\begingroup$ Hi, I edited and linked a photo of my attempt at this problem. I used the comparison test to conclude that since b converge, a must also converge. But I got this problem wrong. I'm not sure why. Thank you! $\endgroup$ Mar 17, 2019 at 3:34
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    $\begingroup$ $ne^n+1<ne^n+e^n$ $\endgroup$
    – Nosrati
    Mar 17, 2019 at 3:44
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    $\begingroup$ $a_n$ will diverge if $b_n$ diverges, because $a_n>b_n$. For proving the convergence, you need to find convergent $c_n$ such that $a_n\le c_n$. $\endgroup$ Mar 17, 2019 at 4:09
  • $\begingroup$ @AnnaNguyen You got the problem wrong because it doesn't help you to show that a series is bigger than a convergent series. You have to show either that it's smaller than a convergent series (in which case it converges) or that it's bigger than a divergent series (in which case it diverges). Bigger than a convergent series doesn't help you at all, and neither does smaller than a divergent series. $\endgroup$ Mar 17, 2019 at 9:26

3 Answers 3

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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).

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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?

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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.

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