Convergence of $\sum_{n=1}^{\infty}\frac{\sqrt n}{2^n}$ In order to find if the series is convergente or divergent:
$$\sum_{n=1}^{\infty}\frac{\sqrt n}{2^n}$$
I did the ratio test: $$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$$
I did:
$$\lim_{n\to \infty}\left|\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{\sqrt{n}}{2^n}}\right|= 2 > 1 $$therefore by the ratio test it should diverge, but this series converge as wolfram alpha says
 A: Observe that $$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to \infty}\left|\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{\sqrt{n}}{2^n}}\right|= \lim_{n\to \infty}\left|\frac{\sqrt{n+1}}{2^{n+1}}\times\frac{2^n}{\sqrt{n}}\right|=\lim_{n\to \infty}\left|\frac{2^n}{2^{n+1}}\right|\times\lim_{n\to \infty}\left|\frac{\sqrt{n+1}}{\sqrt{n}}\right|=\frac12$$
A: As noted in the comments, this series converges by the ratio test. Here we have $a_n=\dfrac{\sqrt n}{2^n}$ so that
$$
\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert
= \lim_{n\to\infty}\left\lvert\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{\sqrt n}{2^n}}\right\rvert
=\lim_{n\to\infty}\frac{2^n}{2^{n+1}}\frac{\sqrt{n+1}}{\sqrt n}
=\frac{1}{2}\lim_{n\to\infty}\sqrt{\frac{n+1}{n}}
=\frac{1}{2}\cdot 1
=\frac{1}{2}
$$
It's very likely you made a typo while carrying out your original computation!
A: You did the limit wrong:
$$\lim_{n\to \infty}\left|\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{\sqrt{n}}{2^n}}\right|= \lim\left|\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{2\sqrt{n}}{2^{n+1}}}\right|=\lim \left|\frac{\sqrt{n+1}}{2\sqrt n}\right|\to \frac{1}{2} < 1$$
A: You can actually show convergence by a different method (which also incidentally includes the ratio test). Note that $\sqrt(n)$ < $n$, so $\sum \sqrt(n)/(2^n)$ < $\sum n/(2^n)$ from n = 1 to $\infty$. The latter sum converges due to the ratio test criterion, so the original sum must converge.
