When is $\sum_{N=1}^{\infty}\exp\left(\ln\left(\frac{\sqrt{N}\ln(N)}{g(N)}\right)-\frac{(g(N))^2}{\ln(N)}\right)<\infty$? Let
\begin{align}
\sum_{N=1}^{\infty}\exp\left(\ln \left(\frac{\sqrt{N}\ln(N)}{g(N)}\right)-\frac{(g(N))^2}{\ln(N)}\right) 
\end{align}
Question:
Let $g(N)=a(\ln(N))^{t}$ where $a \geq 0$ is some constant of your choice. I am interested to know what is the smallest $t$ such that the above expression is finite. Is there a $t$ and $g(N)=a(\ln(\ln(N)))^{t}$ such that the above expression is finite? How do I go about solving such a problem?
My attempt: I know $\sum_{N=1}^{\infty}\frac{1}{N^S}< \infty$ for $S>1$ so I want to manipulate the above expression into this form. Let $g(N)=a\ln(N)$. Then
\begin{align}
\sum_{N=1}^{\infty}\exp\left(\ln \left(\frac{\sqrt{N}\ln(N)}{g(N)}\right)-\frac{(g(N))^2}{\ln(N)}\right) &=\sum_{N=1}^{\infty}\exp\left(\frac{1}{2}\ln (N)-\ln(a)-a^2\ln(N)\right) \\
&=a^{-1}\sum_{N=1}^{\infty} \frac{1}{N^{a^2-1/2}}
\end{align}
and the above expression is finite for any $a>\sqrt{3/2}$.
Edit: I'd like to add one more function. What about $g(N)=aN^{t}$?
My attempt:
\begin{align}
\sum_{N=1}^{\infty}\exp\left(\ln \left(\frac{\sqrt{N}\ln(N)}{g(N)}\right)-\frac{(g(N))^2}{\ln(N)}\right) &=\sum_{N=1}^{\infty}
\exp\left((\frac{1}{2}-t)\ln(N)+\ln(\ln (N))-\ln(a)-a^2\frac{N^{2t}}{\ln(N)}\right) \\
&=a^{-1}\sum_{N=1}^{\infty} \frac{\ln(N)}{N^{t-1/2}}\exp\left(-a^2\frac{N^{2t}}{\ln(N)}\right) 
\end{align}
Now I think that if $t>1/2$ then this is finite because the exponential term will kill everything but not sure how to deal with $t<1/2$.
 A: If $g(N)=a(\ln(\ln(N)))^{t}$, rewrite terms in decreasing asymptotic magnitude and get rid of the insignificant ones.
$\displaystyle \ln \left(\frac{\sqrt{n}\ln(n)}{g(n)}\right)=\frac 12 \ln n+\ln \ln n -t\ln \ln \ln - \ln a$
$\displaystyle \frac{(g(n))^2}{\ln(n)}= \frac{a^2(\ln\ln n)^{2t}}{\ln n}=o(1)$
Hence $$\exp\left(\ln \left(\frac{\sqrt{n}\ln(n)}{g(n)}\right) - \frac{(g(n))^2}{\ln(n)}\right)=\exp\left( \frac 12 \ln n+\ln \ln n -t\ln \ln \ln - \ln a+o(1)\right)=\frac{\sqrt{n}\ln n}{(\ln\ln n)^t}\left(\frac 1a +o(1)\right) \to \infty$$
The series always diverges.
A: First of all $a=0$ is a divide-by-zero case, so it is meaningless. We will consider $a > 0$.
Using some properties of logarithms, you can rewrite everything (in the case $g(N)=a (\ln N)^t$) as
$$a^{-1}\sum_{N=1}^{\infty} \frac{\sqrt{N}(\ln N)^{1-t}}{N^{(a (\ln N)^{t-1})^2}}$$
You can see that for $t < 1$ the $N$-th terms diverges to $\infty$, so we can restrict ourselves to the case $t \ge 1$. As you have already noticed, in the case $a > \sqrt{3/2}$, the minimal $t$ for convergence is indeed $t=1$.
What about $0 < a \le \sqrt{3/2}$? In such a case, you can see that for any $t > 1$ the series is convergent. This is because
$$\lim_{N \to \infty} (a (\ln N)^{t-1})^2 = +\infty$$
so that for $N$ large enough your series has terms bounded above by
$$\frac{\sqrt{N}(\ln N)^{1-t}}{N^2}$$
and hence it is convergent. Hence, there is no minimum $t$, but every $t > 1$ will work.
In the case $g(N)=a (\ln \ln N)^t$, the general $N$-th term diverges to $\infty$ for all $a,t$: you will never achieve convergence.
