0
$\begingroup$

Show that if $\sum_{n=1}^\infty a_n$ converges, then $\sum_{n=1}^\infty \left(\frac{1+\sin(a_n)}{2}\right)^n$ converges.

I tried to prove it using the Comparison Test and the Ratio Test but I am not able to come to a conclusive result using those two. I've tried all other convergence tests available to me but I have no luck solving this question. Any guidance would be greatly appreciated.

$\endgroup$
1
  • $\begingroup$ Note that $\lim\limits_{n\rightarrow\infty}a_n=0$, so for big $n$ we have $|a_n|<\frac{1}{2}$, so $|\frac{1+\sin a_n}{2}|\leq \frac{3}{4}$. Does it help? $\endgroup$
    – richrow
    Feb 27, 2020 at 18:43

2 Answers 2

0
$\begingroup$

Hint: $|a_n| \to 0$, so for sufficiently large $n$ you'll have $|1+\sin(a_n)|/2 < 3/4$.

$\endgroup$
0
$\begingroup$

The idea will be if $\sum a_n < \infty$ then $a_n \to 0$, which would mean $\sin a_n \to 0$ as well, since for $x \to 0$, you have $\sin x \approx x$. So for $n > N$, each term in your series can be bounded by $$ \frac{1 + \sin(a_n)}{2} < \frac{2}{3}, $$ or any other number in $(1/2,1)$ and now your sum bounds by geometric series.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.