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I'm stuck on this.. Given $\sum_{n=1}^{\infty}\frac{n!}{n^n}\cdot sin(n^2)$ I have to determine if it's convergent or not.

I can see that $sin(n^2)$ is bounded and so are the partial sums of it. But $\frac{n!}{n^n}$ doesnt converge to $0$ to use Dirichlet's test

Also my intuition is that it's not convergence but I can't find another divergent sequence to use the comparison test

Any hints?

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    $\begingroup$ $n!/n^n$ does tend to $0$, and hence your serie converges absolutely, and hence converges $\endgroup$ – Thinking Jan 28 '19 at 19:03
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    $\begingroup$ By the way, partial sums of $\sin n^2$ are not bounded. $\endgroup$ – RRL Jan 28 '19 at 19:07
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Hint $$\frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot...\cdot\frac{n}{n}\leq \frac{1}{n}\cdot\frac{2}{n}\cdot 1 \cdot 1 \cdot... \cdot 1=\frac{2}{n^2}$$

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  • $\begingroup$ So if $\frac{n!}{n^n} \leq \frac{2}{n^2}$ that means that $\frac{n!}{n^n} \leq \frac{1}{n}$ and $\frac{n!}{n^n} \rightarrow 0$. So I can use Dirichlet's test? Is that right? $\endgroup$ – VakiPitsi Jan 28 '19 at 19:19
  • $\begingroup$ No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $\left | sin(x^2) \right | \leq1\Rightarrow \left |\frac{n!}{n^n}sin(x^2) \right |\leq \frac{1}{n^2}$ and thus the series converges right? $\endgroup$ – VakiPitsi Jan 28 '19 at 19:25
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    $\begingroup$ @VakiPitsi Exactly. You should maybe emphasize that you are proving absolute convergence, hence convergence. $\endgroup$ – N. S. Jan 28 '19 at 19:47
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    $\begingroup$ Can the downvoter please explain what is wrong with the hint. $\endgroup$ – N. S. Jan 28 '19 at 19:48
  • $\begingroup$ I wasn't the downvoter, but did you mean $\frac 2 {n} . 1$ rather than $\frac 1 {n} . 1$ ? $\endgroup$ – J. W. Tanner Jan 28 '19 at 20:00
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Hint: if $k$ is the integer part of $n/2$, then $$\frac{n!}{n^n} \leq \frac{k!}{n^k} \leq (k/n)^k \leq 2^{-k}.$$

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