# prove $\sum_{n=1}^\infty \frac{1}{2^\sqrt{n}}$ converges

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

converges

Tools I have to use: comparison test, limit comparison test, root test, ratio test.

What I have tried:

I claimed that $\dfrac{1}{2^\sqrt{n}} < \dfrac{1}{n^2}$ for some $n$ say $n = k$ which is true as exponential beats power, then used the comparison test. However this does not seem sleek enough and I feel as though I'm missing something obvious

• Your method looks sleek enough to me, but you have to say more than "exponential beats power" to justify it. Also your claim should be that there exists $k$ such that $\frac{1}{2^{\sqrt n}} < \frac{1}{n^2}$ for all $n \ge k$. – TonyK Apr 26 '14 at 22:45
• look at answers of this similar question. – achille hui Apr 26 '14 at 22:45
• @TonyK Yes that's what I claimed, but I can't seem to solve for a particular n, so I don't know how to prove that claim - so I just stated it – user144464 Apr 26 '14 at 22:55
• Have you tried the integral test? – mjh Apr 26 '14 at 23:13

$$g(x) = \frac{x^2}{2^{\sqrt{x}}} \quad\implies\quad g'(x) = \left(2 - \frac{\log 2}{2} \sqrt{x}\right)\frac{x}{2^{\sqrt{x}}}$$
When $x > \left(\frac{4}{\log 2}\right)^2 \sim 33.3019$, $g'(x) < 0$ and hence monotonic decreasing. This means if you can find a $N \ge 34$ such that $g(N) \le 1$, then for all $n \ge N$, we will have
$$g(n) \le g(N) \le 1\quad\implies\quad \frac{1}{2^{\sqrt{n}}} \le \frac{1}{n^2}$$ If one try $n$ of the form $2^\alpha$, one find $g(256) = g(2^8) = 1$. i.e. for all $n \ge 256$, we have $\frac{1}{2^{\sqrt{n}}} \le \frac{1}{n^2}$.
For other ways to attack this sort of problem, look at answers of a similar question which asks for the convergence of $\sum_{n=1}^\infty\frac{1}{3^{\sqrt{n}}}$.