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Show that $$\sum_{k=1}^\infty \log(k+1) \frac{e^{\sin(k)}}{e^{k}}$$ converges.

Now, here's the thing. I have taken this exercise from a textbook, and this textbook wants you to do this without the ratio test, because the ratio test hasn't been introduced yet in this book up to this point. Essentially all I have is the alternating series test, which doesn't seem to be applicable here, and direct comparison test (and of course all of the convergence tests and rules for real sequences in general, like if it's monotone and bounded or if it's a Cauchy sequence)

Any hints?

EDIT: Attempted solution with hints:

I know that $k^3 < e^k$ for all $k > 4 \cdot 3^2 = 36$. Then, if $k > 37$, we have $k^3 < e^{k-1}$, so we also have $$\frac{k}{e^{k-1}} < \frac{1}{k^2}$$

Then, by the direct comparison test $$\sum_{k=1}^\infty \frac{k}{e^{k-1}}$$ converges, as $\sum_{k=1}^\infty \frac{1}{k^2}$ converges. Then, since $\log(k+1) < k$, we also have $$\frac{\log(k+1)}{e^{k-1}} < \frac{k}{e^{k-1}}$$

Then, by the direct comparison test $$\sum_{k=1}^\infty \frac{\log(k+1)}{e^{k-1}}$$ converges as well. Then, since $e^{\sin(k)} \leq e$, we also have

$$\frac{\log(k+1)}{e^{k}} \cdot e^{\sin(k)} \leq \frac{\log(k+1)}{e^{k}} \cdot e$$

which is equivalent to

$$\frac{\log(k+1)}{e^{k}} \cdot e^{\sin(k)} \leq \frac{\log(k+1)}{e^{k-1}}$$

Then, by the direct comparison test

$$\sum_{k=1}^\infty \frac{\log(k+1)}{e^{k}} \cdot e^{\sin(k)} $$

converges.

Is this (somewhat) correct?

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  • $\begingroup$ Oops: the textbook wants you to do this without the ratio test... $\endgroup$
    – amWhy
    Jul 9 '18 at 21:16
  • $\begingroup$ @amWhy I have been thinking about this for 2 hours, but without any further hints I don't think I'll be able to do this... It even took me a considerable amount of time to prove that the terms of the series are a null sequence $\endgroup$
    – user295213
    Jul 9 '18 at 21:17
  • $\begingroup$ I'm simply trying to establish who "you" is in this case. You've included background (what and what you haven't covered), stated the problem, and are asking only for hints, so I have not problem whatsoever with your question. $\endgroup$
    – amWhy
    Jul 9 '18 at 21:22
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HINTS:

For $k\ge 1$, we have

$$0< e^{\sin(k)}\le e$$

and

$$0<\log(1+k)\le k$$

and

$$\frac16 k^3<e^k$$

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  • $\begingroup$ I used your hints to come up with a solution. Can you glance over it to see if it's correct? $\endgroup$
    – user295213
    Jul 10 '18 at 0:51
  • $\begingroup$ Your reasoning looks fine. $\endgroup$
    – Mark Viola
    Jul 10 '18 at 1:09
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hint

Use the fact that $$e^{\sin( k)}\le e$$ and

$$\lim_{k\to+\infty}k^2\ln(k+1)e^{-k}=0.$$

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  • $\begingroup$ Thanks for the hints! $\endgroup$
    – amWhy
    Jul 9 '18 at 21:26
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Simply $\;\log(k+1)\dfrac{\mathrm e^{\sin k}}{\mathrm e^k}=o\biggl(\dfrac1{\mathrm e^{k/2}}\biggr)$ since $\;\lim_{k\to\infty}\dfrac{\log(k+1)\,\mathrm e^{\sin k}}{\mathrm e^{k/2}}=0$, and the latter is a geometric series with ratio $\dfrac1{\sqrt e}<1$.

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