0
$\begingroup$

Let $n \rightarrow \infty$ and consider

$$\sum_{x=\lfloor \log^6(n)\rfloor}^{\lceil \frac{n}{\log^2(n)}\rceil} \left(\frac{n}{\log^2(n) x} e^{-\frac{\log^{16}(n)}{n}}\right)^x$$

Do we know anything about the convergence of this sum? In particular, does it maybe go to 0?

Thank you very much for any hints!

$\endgroup$
  • 1
    $\begingroup$ The summation notation only really makes sense when the bounds are integers. $\endgroup$ – T.J. Gaffney May 10 '16 at 18:30
  • $\begingroup$ Thanks for the remark. Let the bounds be floored and ceiled, I will adapt this in the question. $\endgroup$ – user136457 May 10 '16 at 18:31
  • 1
    $\begingroup$ Technically true, but throw in some floors and ceiling in the mix -- it won't change the convergence. @Gaffney $\endgroup$ – Clement C. May 10 '16 at 18:31
1
$\begingroup$

You can lower bound your sum $S_n$ by its first term, $$ S_n \geq \left( \frac{n}{\log^8 n} e^{-\frac{\log^{16} n}{n}} \right)^{\log^6 n} = e^{\log^7 n - 8\log^6 n \cdot \log\log n - \frac{\log^{22} n}{n}} = e^{\log^7 n + o(\log^7 n)} \xrightarrow[n\to\infty]{} \infty $$

$\endgroup$
  • $\begingroup$ Wow, such an easy and nice solution. Thank you very much for that! $\endgroup$ – user136457 May 10 '16 at 18:39

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.