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We have the following infinite sum and must know whether it absolutely or conditionally converges or diverges

$$\sum_{i=1}^{\infty} (-1)^n\frac{e^{1/n}}{n^3}$$

I first proved it conditionally converges by the alternating series test, however my problem is proving if it absolutely converges.

I started off by taking the absolute value which gives

$$\sum_{i=1}^{\infty} \frac{e^{1/n}}{n^3}$$

I first tried the ratio test but the limit of the ratio of consecutive terms gives a value of 1, therefore inconclusive. I next tried the integral test, but as most of you may know $e^{1/x}$ has no elementary function which makes this test useless. Next i tried the limit comparison test with the general term $\frac{e^n}{n^3}$, but in order for that to work we need to prove that the general term we are comparing with converges, which i also had problems with(ratio test is inconclusive aswell and could not come up with a function to compare it with).

I am at a dead end and would like a hint towards proving this infinite series absolutely converges.

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    $\begingroup$ $\frac{e^{\frac{1}{n}}}{n^3}$ is asymptotically equivalent to $\frac{1}{n^3}$, since $$\lim_{n \to \infty} \frac{e^{\frac{1}{n}}}{n^3} \cdot n^3=e^0=1$$ so $$\sum \frac{e^{\frac{1}{n}}}{n^3} < \infty $$ $\endgroup$ – Chinnapparaj R Apr 19 at 9:16
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$|(-1)^{n}e^{1/n}| \leq e^{1}=e$. Compare with $\sum \frac 1 {n^{3}}$.

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  • $\begingroup$ I see, so because the denominator dictates the behavior of the expression $\frac{e^{1/n}}{n^3}$ we must find an expression that also has a denominator that dictates the behavior of the expression very similarly regardless of the numerator in order to do the limit comparison test? $\endgroup$ – Not Friedrich gauss Apr 19 at 9:27

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