Is $\sum_{n=1}^{\infty}\frac{\sin a_{n}}{\sqrt{n}+na_{n}}$ convergent?

Question

Suppose $a_{n}>0$ and $\sum_{n=1}^{\infty}a_{n}$ is convergent.

$\sum_{n=1}^{\infty}\frac{\sin a_{n}}{\sqrt{n}+na_{n}}$ convergent ?

Since $\sin a_{n}$ is bounded by one,

and

$\sqrt{n}\rightarrow\infty$ as $n\rightarrow\infty$, so $\frac{\sin a_{n}}{\sqrt{n}+na_{n}}\rightarrow0$.

I tried limit comparison test,

for instance $\sum_{n=1}^{\infty}\frac{1}{n}$, $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$, and $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$, but did work. Any hint?

Thank you very much.

Answer 1

$$\vert b_n \vert= \vert \frac{\sin a_{n}}{\sqrt{n}+na_{n}} \vert \leq \vert \frac{\sin a_n}{\sqrt{n}} \vert \leq \vert \frac{a_n}{\sqrt{n}} \vert \leq \vert a_n \vert $$ where the first inequality is due to the fact that $a_n \geq 0$ and the second one is due to $\sin x < x $.

Now you can say by Comparison test that $\vert b_n \vert$ converges. Hence, $b_n$ is absolutely convergent.

Thanks @nosrati for the comment.

Answer 2

$$\frac{\sin a_{n}}{\sqrt{n}+na_{n}}\le\frac{\sin a_{n}}{\sqrt{n}+\sqrt{n} a_{n}}\le\frac{\sin a_{n}}{\sqrt{n}\cdot (1+a_{n}) }\le\left|\frac{a_n}{\sqrt{n}}\right| $$

$$\frac{a_{n+1}\sqrt{n}}{a_n\sqrt{n+1}}\le \frac{a_{n+1}}{a_n}<1$$

as we know that $1+a_n\ge0$

Answer 3

$$ b_n = \frac{\sin a_n}{\sqrt{n} + n a_n} = \frac{a_n}{\sqrt{n}}\frac{\sin a_n}{a_n}\frac{1}{1+\sqrt{n}a_n} $$ $|(\sin a_n)/a_n|\le 1$, and since the $a_n$ are all positive, $|1+\sqrt{n} a_n|^{-1} \le 1$. This means we have $$ |b_n| \le \frac{a_n}{\sqrt{n}}\le a_n $$ for all $n$. Since $\sum_{n=0}^\infty a_n$ converges, by the comparison test $\sum_{n=0}^\infty b_n$ also converges.

Answer 4

Since $|\sin x| \le |x|$ for all $x,$ we have

$$\left|\frac{\sin a_n}{\sqrt n +na_n}\right |= \frac{|\sin a_n| }{\sqrt n +na_n} \le \frac{|a_n| }{\sqrt n +na_n}$$ $$ = \frac{a_n }{\sqrt n +na_n} \le a_n$$

for all $n.$ Because $\sum a_n < \infty,$ our series converges absolutely by the comparison test. Hence it converges.