How to bound $\sum_{i=2}^{i_0}e^{2i}/i^2$? I am reading an handout and it indicates the following inequality. And I want to know why it is true.
$$\sum_{i=2}^{i_0-1}e^{2i}/i^2\le 8(\frac{n}{\log n})(\log n)^{-2},$$
where $i_0=(\log n-\log\log n)/2$ and we can assume $n$ is large enough. And the exact value 8 is not important for me.
My strategy is $\sum_{i=2}^{i_0-1}e^{2i}/i^2\le e^{2i_0}\sum 1/i^2$, but it will lose the factor $(\log n)^{-2}$.
 A: Because $\frac{e^{2i}}{i^2}$ is increasing function we can get an upper bound by :
$\sum \limits_{i=2}^{i_0-1} \frac{e^{2i}}{i^2} \leq \int \limits_{2}^{i_0} \frac{e^{2t}}{t^2} dt =2 \text{li}\left(e^{2 t}\right)-\frac{e^{2 t}}{t}|_{2}^{i_0} = 2 \text{li}\left(e^{2 i_0}\right)-\frac{e^{2 i_0}}{i_0} +\approx 11.9627   <c_1 (2 li(e^{2i_0})-\frac{e^{2i_0}}{i_0}) $ for some $c_1 >1$ 
So we arrive at $c_1 (2 li(e^{2i_0})-\frac{e^{2i_0}}{i_0})$ such that $i_0 = \frac{\ln n -\ln \ln n}{2}$, by substituting that in the function above we get :
$c_1(2 \text{li}\left(\frac{n}{\log (n)}\right)-\frac{2 n}{\log ^2(n)-\log (n) \log
   (\log (n))})$ and since $li(n) \leq \frac{n}{\ln n}(1+\frac{2}{\ln n})$ so the function above is less than :
$c_1 \frac{2 n \left(\frac{\log \left(\frac{n}{\log (n)}\right)+2}{\log
   ^2\left(\frac{n}{\log (n)}\right)}+\frac{1}{\log (\log (n))-\log
   (n)}\right)}{\log (n)}$ now we can see the first two terms the coefficient term and the $\frac{n}{\ln n}$ term which becomes :
$2c_1 \frac{n}{\ln n} \left( \frac{\log \left(\frac{n}{\log (n)}\right)+2}{\log ^2\left(\frac{n}{\log
   (n)}\right)}+\frac{1}{\log (\log (n))-\log (n)}\right) $ we need to prove that $\frac{c_2}{\ln^n} \leq \frac{\log \left(\frac{n}{\log (n)}\right)+2}{\log ^2\left(\frac{n}{\log
   (n)}\right)}+\frac{1}{\log (\log (n))-\log (n)} \leq \frac{c_3}{\ln^2 n}$
by simple logarithmic proprieties and laws we arrive at :
$\frac{\log \left(\frac{n}{\log (n)}\right)+2}{\log ^2\left(\frac{n}{\log
   (n)}\right)}+\frac{1}{\log (\log (n))-\log (n)}=\frac{1}{\log (\log (n))-\log
   (n)}+\frac{1}{\log (n)-\log (\log (n))}+\frac{2}{(\log (n)-\log (\log (n)))^2}$, now combine the first two terms we get $0$ since $\frac{1}{x}+\frac{1}{-x} = 0$ for all $x \not= 0$.
so we get the final result $\frac{2}{\ln^2 n} \leq \frac{2}{(\log (n)-\log (\log (n)))^2} \leq \frac{c_3}{\ln^2 n}$ such that $c_3 > 2$.
so $max(c_2=2 ,c_3 >2) = c_3$ so we get that :
$\sum \limits_{i=2}^{i_0-1} \frac{e^{2i}}{i^2} \leq 2 c_1 c_3 \frac{n}{\ln n} \frac{1}{\ln ^2 n}$ because $c_1 >1$ and $c_3 >2$ the coefficient must be bigger than $2*1*2=4$ so $8$ is fine,which proves the inequality.
Note : i think the writer used $8$ to make the inequality true for all numbers $n$. because the proof above is valid for all $n\geq n_0$ such that $n_0$ is defined by the coefficient.
A: We can separate the sum into 2 parts: $i\le (\log n-3\log\log n)/2$ and others. Then bound the first part by the method I mentioned in the question, and the second part by sum of geometric series.
