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I am very confident that the ratio test can be used here so I might have a mistake in the calculation

$$\sum_{k=1}^{\infty} \frac{k^{100}}{e^k}$$

Applying the ratio test this becomes:

$$\frac{\frac{(k+1)^{100}}{e^{k+1}}}{\frac{k^{100}}{e^k}}$$

which transforms to

$$\frac{(k+1)^{100}}{e^{k+1}}\frac{e^k}{k^{100}}$$

thus

$$\frac{(k+1)^{100}}{ek^{100}}$$

I am unable to proceed any further

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    $\begingroup$ Now do $=\frac{1}{e}((k+1)/k)^{100}=\frac{1}{e}(1+1/k)^{100}\to\frac{1}{e}$. $\endgroup$ – conditionalMethod Dec 4 '19 at 17:50
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You have that $$ \begin{gathered} \mathop {\lim }\limits_{k \to + \infty } \frac{{(k + 1)^{100} }} {{ek^{100} }} = \frac{1} {e}\mathop {\lim }\limits_{k \to + \infty } \left( {\frac{{k + 1}} {k}} \right)^{100} = \hfill \\ \hfill \\ = \frac{1} {e}\mathop {\lim }\limits_{k \to + \infty } \left( {\frac{{k + 1}} {k}} \right)^{100} = \frac{1} {e} \cdot 1^{100} = \frac{1} {e} < 1 \hfill \\ \end{gathered} $$ therefore your series is convergent

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    $\begingroup$ Exactly, indeed I also think that the confusion is on that step $$\frac{k+1}{k} \to 1 \implies \left(\frac{k+1}{k}\right)^{n} \to 1^n=1 \quad \forall n$$ $\endgroup$ – user Dec 4 '19 at 17:56
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Your way is right, to conclude we simply have

$$\frac{(k+1)^{100}}{ek^{100}}=\frac1e \left(\frac{k+1}{k}\right)^{100}\to \frac1e \cdot 1^{100}=\frac1e <1$$

indeed

$$\frac{k+1}{k}=1+\frac1k \to 1 \implies \left(\frac{k+1}{k}\right)^{n} \to 1^n=1 \quad \forall n$$

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