Prove that the series converges and find an upper bound on $\alpha$.
$\alpha$:= $\sum_{n=10^{10}}^{\infty} \frac{1}{nlog_{10}(n)log_{10}(log_{10}(n))^3}$
This is my work: http://imgur.com/a/3K1vx
I'm really bad with logarithmic functions so I'm not really sure how I did, would someone mind checking my work please? Also, what would the upper bound be and how do you find that?
For $n$ between $10^k$ and $10^{k+1}$ you get $$ \frac1{10^{k+1}·(k+1)·[\log_{10}(k+1)]^3}\le a_n\le \frac1{10^k·k·[\log_{10}(k)]^3} $$ There are $10^{k+1}-10^k=9·10^k$ elements in that index segment if you exclude one of the endpoints. Summation gives thus $$ \frac{9·10^k}{10^{k+1}·(k+1)·[\log_{10}(k+1)]^3} \le\sum_{n=10^k+1}^{10^{k+1}} a_n \le \sum_{n=10^k}^{10^{k+1}-1} a_n \le\frac{9·10^k}{10^k·k·[\log_{10}(k)]^3} $$ Thus $$ \frac1{10^{11}}+\frac{9}{10}\sum_{k=11}^\infty \frac1{k[\log_{10}(k)]^3}\le α \le 9\sum_{k=10}^\infty \frac{1}{k[\log_{10}(k)]^3} $$ Then condensate once more. $$ \frac9{10}\sum_{\ell=2}^\infty\frac1{\ell^3} \le\sum_{k=10^1+1}^\infty \frac{1}{k[\log_{10}(k)]^3},\quad \sum_{k=10^1}^\infty \frac{1}{k[\log_{10}(k)]^3} \le 9\sum_{\ell=1}^\infty\frac1{\ell^3} $$
In total you get for the upper bound \begin{align} α&\le\sum_{k=10}^\infty\frac{9·10^k}{10^k·k·[\log_{10}(k)]^3} =9·\sum_{k=10}^\infty \frac{1}{k[\log_{10}(k)]^3}\\ &\le9\sum_{\ell=1}^\infty\frac{9·10^\ell}{10^\ell·\ell^3} =81·\sum_{\ell=1}^\infty\frac{1}{\ell^3}\\ &\le 81·\left(1+\sum_{\ell=2}^\infty\frac{1}{\ell^3-\ell}\right) =81·\left(1+\frac12\lim_{L\to\infty}\left(\frac1{1·2}-\frac1{L(L+1)}\right)\right)=\frac{405}4 \end{align}