1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $n \ge k$. $\endgroup$ – TonyK Apr 26 '14 at 22:45
  • $\begingroup$ look at answers of this similar question. $\endgroup$ – achille hui Apr 26 '14 at 22:45
  • $\begingroup$ @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 $\endgroup$ – user144464 Apr 26 '14 at 22:55
  • $\begingroup$ Have you tried the integral test? $\endgroup$ – mjh Apr 26 '14 at 23:13
1
$\begingroup$

Consider the function

$$g(x) = \frac{x^2}{2^{\sqrt{x}}} \quad\implies\quad g'(x) = \left(2 - \frac{\log 2}{2} \sqrt{x}\right)\frac{x}{2^{\sqrt{x}}}$$

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 $\sum_{n=1}^\infty\frac{1}{3^{\sqrt{n}}}$.

$\endgroup$

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.