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I'm currently working through a homework problem, but I've been stumped on this one problem for almost a day now. I don't want the problem to be worked out, rather I just want some idea of where to start/if my thought process has been incorrect.

The problem is to figure out whether or not the following series satisfies the M-test and converges uniformly on the interval $[0,1]$.

$$\sum_{k=1}^\infty \frac{e^{x/k}-1}{k}$$

I initially tried to see if $$\sum_{k=1}^\infty \frac{e^{1/k}-1}{k}$$ converges, since $$\frac{e^{x/k}-1}{k} \le \frac{e^{1/k}-1}{k}$$ when $x \in [0,1]$. However, after trying most of the convergence tests (Integral, comparison, ratio, root, etc) I wasn't able to conclude if this bound converges or not.

However, I've also been unable to show if $\sum_{k=1}^\infty \frac{e^{x/k}-1}{k}$ diverges, which would mean that it would fail the M-test.

Does anybody have any suggestions as to where to start with this problem?

Melon
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    $e^{t}=1+t+o(t)$ as $ t \to 0$. – geetha290krm Feb 19 '23 at 04:55
  • Oh! Would I express the numerator $e^{1/k}-1$ as $x/k + o(x^2/k^2)$?

    I think then I would have that each term would be $1/k^2 + o(x^2/k^3)$ which is the sum of two convergent series by the P-series test, right?

     – Melon Feb 19 '23 at 05:01
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    Why are you worrying about $x/k$? Just concentrate on the right-hand side, since it gives you a uniform bound on $[0,1]$. Does $\sum (e^{1/k}-1)/k$ converge? By the way, your list of convergence tests is missing what I think is the most powerful test — the limit comparison test. – Ted Shifrin Feb 19 '23 at 05:13
  • Oh thats my bad. I don't know why I typed x/k, as I was looking at the right-hand side already. I meant to say that maybe I could expand $e^{1/k}-1$ as $1/k^2 + o(1/k^2)$ since I was trying to show that $\sum (e^{1/k}-1)/k$ converges for the M-test. Thanks for the advice though, the limit comparison test wasn't taught in my class, but I have a feeling I should have already known this from previous courses... – Melon Feb 19 '23 at 08:12

1 Answers1

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$$\sum_{k=1}^\infty\frac{e^{1/k}-1}k=\sum_{k=1}^\infty\sum_{j=1}^\infty\frac1{k^{j+1}j!}$$ $$=\sum_{j=1}^\infty\frac1{j!}\sum_{k=1}^\infty\frac1{k^{j+1}}=\sum_{j=1}^\infty\frac{\zeta(j+1)}{j!}$$ $$<\sum_{j=1}^\infty\frac2{j!}=2(e-1)$$ Therefore the bound converges and the first series converges uniformly.

Parcly Taxel
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    Solved more simply in the duplicate, with the limit comparison test, as advised by Ted Shifrin in the comment above. – Anne Bauval Feb 19 '23 at 06:32