# Is the series $\sum_{n=1}^\infty \frac{n^2}{2^{\sqrt n}}$ convergent?

Is this series convergent? $$\sum_{n=1}^\infty \frac{n^2}{2^{\sqrt n}}$$

Edit: I'm sorry I wrote a double negative. This is the series. (thanks Brian M. Scott for editing)

• Go to the FAQ section adn read the directions there to use LaTeX in this site. – DonAntonio Feb 25 '13 at 19:33
• Probably you do not mean a negative exponent at the bottom. – André Nicolas Feb 25 '13 at 19:34
• Here is one place to start learning how to write mathematics here. – Brian M. Scott Feb 25 '13 at 19:34
• If the exponent downstairs is negative, the terms grow without bound. So no. – vonbrand Feb 25 '13 at 19:35

Note that $2^{\sqrt{n}}=e^{(\log 2)\sqrt{n}}$.

By looking at the power series expansion of $e^x$, we can see that for positive $x$, we have $e^x\gt \frac{x^8}{8!}$. Thus $$e^{\log 2\sqrt{n}}\gt \frac{(\log 2)^8}{8!}n^4.$$

So the $n$-th term of our sequence is $\lt \frac{8!}{(\log 2)^8}\frac{1}{n^2}$. Comparison with the convergent series $\sum\frac{1}{n^2}$ shows that our series converges.

• Really nice answer! – Jonathan H Feb 25 '13 at 19:55
• Wow thank you great answer! It actually helped me with some other series I had problems with! – milo Feb 25 '13 at 20:06
• You are welcome. There are easier ways, since there is lots of slack, the terms go down really fast. But I have a fondness for explicit (even if weak) bounds. – André Nicolas Feb 25 '13 at 20:11

You wrote

$$\sum_{n=1}^\infty \frac{n^2}{2^{-\sqrt n}}=\sum_{n=1}^\infty 2^{\sqrt n}n^2$$

But this series clearly diverges since $\,2^{\sqrt n}n^2\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{} 0\,$

No, it's not convergent, since the $n^2/2^{-\sqrt n}\to\infty$ for $n\to\infty$. But $$\sum_{n=1}^\infty\frac{n^2}{2^{\sqrt n}}$$ converges, because the summands go to zero faster than $1/n^2$ for large enough $n$.

$$\frac{n^2}{2^{-\sqrt{n}}}={n^2}{2^{\sqrt{n}}}$$which does not converge to $0$. Therefore, the series cannot converge.