Does $\sum_{k\geq 1} e^{-\frac{k^{q}}{\sqrt{logk}}}<\infty$, where $0<q<1?$ Does $\sum_{k\geq 1} e^{-\frac{k^{q}}{\sqrt{2logk}}}&lt;\infty$, where $0&lt;q&lt;1?$

Or that $\sum_{k\geq 1} (1-\frac{1}{\sqrt{2logk}k^{1-q}} )^{k} &lt;\infty$?

But what I really want to bound is $\sum_{k\geq 1} (1-\frac{\sqrt{2logk}}{(2logk+1) k^{1-q}} )^{k} &lt;\infty$, simply the above two are dominating for this one.

Attempt

1)Ratio and root test give limit equal to 1

2) Trying to get $\int_{1}^{\infty} e^{-\frac{x^{q}}{\sqrt{logx}}}dx$ into some known integral. A good candidate was the Gamma integral $\Gamma(\frac{1}{q})=\frac{1}{q}\int x^{1-\frac{1}{q}} e^{-x^{q}}dx =\int e^{-x^{q}}dx$. But $x^{q}&gt;\frac{x^{q}}{\sqrt{logx}}$ and integration by parts gives 

$$dy = (\frac{x^{q}}{\sqrt{logx}})'dx,$$
which derivative I can't find any write it in terms of y.

3)Say $q=\frac{1}{2}$. I tried to lower bound $\frac{x^{q}}{\sqrt{logx}}$, we have the error term from $e^{x}$'s Taylor expansion

$$\sqrt{\frac{x}{logx}}=(log(\sum_{k\neq 1}\frac{x^{k}}{k!}+1)+1)^{1/2}.$$

By lower bounding maybe we get a converging dominating series.
First of all the series must start from $2$ (you have a log at the denominator). Then note that for all sufficiently small $\epsilon&gt;0$ holds, for a sufficiently large $k$, $k&gt;N$ say, $$\frac{k^{q}}{\sqrt{\log\left(k\right)}}\geq k^{q-\epsilon}$$ since $k^{\epsilon}$ grows faster than $\sqrt{\log\left(k\right)}$. So $$\sum_{k&gt;N}\exp\left(-\frac{k^{q}}{\sqrt{\log\left(k\right)}}\right)\leq\sum_{k&gt;N}\exp\left(-k^{q-\epsilon}\right).$$ Now note that $$ e^{k^{a}}\geq k^{1+a}\Leftrightarrow k^{a}\geq\left(1+a\right)\log\left(k\right)$$ and this is true since every power of $k$ grow faster than $\log(k)$. So if we take $a=q-\epsilon$, for a sufficiently large $k$, we have $$\sum_{k&gt;N}\exp\left(-k^{a}\right)\leq\sum_{k&gt; N}k^{-\left(1+a\right)}\leq\sum_{k\geq1}k^{-\left(1+a\right)}=\zeta\left(1+a\right)$$ so the series is convergent.
small positive powers dominate large logarithmic powers, so k^q > (log(k)) ^ N for k large enough depending on q > 0, N arbitrarily big fixed, so your series is dominated by a series with e^-(log(k)^M) for arbitrarily big fixed M, and in particular for M=2, you get the series 1/(k^log(k)) which is obviously convergent as it is smaller than any zeta (N), N>2 say