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

Question

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?

Answer 1

HINTS:

For $k\ge 1$, we have

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

and

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

and

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

Answer 2

hint

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

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

Answer 3

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 $\dfrac1{\sqrt e}<1$.