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!

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    $\begingroup$ 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
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    $\begingroup$ 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

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

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

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