How to find example for this sequence? i'm currently studing about sums and sequence,  and i've been struggle to find an exmaple for this task :  
find a sequence $a_n$ that approaches to $0$ , that the sum $$\sum _{n=0}^{\infty }\:\frac{1}{n^{\left(1+a_n\right)}}$$ converge
Any ideas?
 A: To make the series converge you want the $a_n$ as large as possible (and positive) while still converging to $0$. In other words, you want a sequence that converges to $0$ as slowly as possible.
I think $a_n = \dfrac{1}{\sqrt{\ln{n}}}$ will work. It obviously converges to $0$ but the hard part is proving series convergence.
I think the Integration Test for convergence would do it. To integrate:
$\dfrac{1}{x^{\left(1+\dfrac{1}{\sqrt{\ln{x}}}\right)}}$ is not simple though.
Convert it to:
\begin{eqnarray*}
\dfrac{1}{x.x^{\left(\dfrac{1}{\sqrt{\ln{x}}}\right)}} &=& \dfrac{1}{x.e^{\ln{\left(x^{\left(\dfrac{1}{\sqrt{\ln{x}}}\right)}\right)}}} \\
&=& \dfrac{1}{x.e^{\left(\dfrac{1}{\sqrt{\ln{x}}}\right)ln{\left(x\right)}}} \\
&=& \dfrac{1}{x.e^{\left(\sqrt{\ln{x}}\right)}} \\\end{eqnarray*}
Integrate by parts with $u = \sqrt{\ln{x}}\qquad$ and $\qquad dv = \dfrac{1}{x\sqrt{\ln{x}}e^{\left(\sqrt{\ln{x}}\right)}}$
So $\qquad du = \dfrac{1}{2x\sqrt{\ln{x}}}\qquad$ and $v = \dfrac{-2}{e^{\sqrt{\ln{x}}}}$.
So the integral is $$-\dfrac{2\sqrt{\ln{x}}}{e^{\sqrt{\ln{x}}}} + \int \dfrac{1}{x\sqrt{\ln{x}}e^{\left(\sqrt{\ln{x}}\right)}} dx $$
$$= -\dfrac{2\sqrt{\ln{x}}}{e^{\sqrt{\ln{x}}}} - \dfrac{2}{e^{\sqrt{\ln{x}}}}$$
Evaluate from $e$ to $\infty$:
$\left[-\dfrac{2\sqrt{\ln{x}}}{e^{\sqrt{\ln{x}}}} - \dfrac{2}{e^{\sqrt{\ln{x}}}}\right]_{e}^{\infty} = -2\left[0 - \dfrac{2}{e}\right] = \dfrac{4}{e}$
That evaluation at $\infty$ assumes $\lim_{x\to\infty}\dfrac{\sqrt{\ln{x}}}{e^{\sqrt{\ln{x}}}} = 0$. We prove it with l'Hopital's rule:
\begin{eqnarray*}
\lim_{x\to\infty}\dfrac{\sqrt{\ln{x}}}{e^{\sqrt{\ln{x}}}} &=& \lim_{x\to\infty}\dfrac{1}{2x\sqrt{\ln{x}}} . \dfrac{-1}{2x\sqrt{\ln{x}}e^{\sqrt{\ln{x}}}} \\
&=& \lim_{x\to\infty} \dfrac{-1}{4x^2\left(\ln{x}\right)e^{\sqrt{\ln{x}}}} \\
&=& 0
\end{eqnarray*}
The integral giving a finite result proves convergence of the series.
