Can some one tell me the asymptotic behavior of following series for large n? $$\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^n \left(\frac{1}{\left(2k-1+2^{1/\alpha}\right)^\alpha} - \frac{1}{\left(2k+2^{1/\alpha}\right)^\alpha}\right),$$ where $0<\alpha<2$.
 A: As I suggested, let us start with the case $\alpha=1$ because it is simplest.  In that case
$$\sum\limits_{k=1}^n \left(\frac{1}{\left(2k-1+2^{1/\alpha}\right)^\alpha} - \frac{1}{\left(2k+2^{1/\alpha}\right)^\alpha}\right)=\sum\limits_{k=1}^n \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right)\\
=\sum\limits_{k=1}^n \left(\frac{1}{(2k+1)(2k+2)} \right)\\=\frac 12 \left(H_{n+\frac 12}-H_{n+1}-1+\log(4)\right)$$
where $H_n$ is the $n$th harmonic number and the calculation is from Alpha.
  As $H_{n+\frac 12}-H_{n+1} \approx \log(n+1)-\log(n+\frac 12)\approx \frac 1{2n}$ the sum is about $\log(2)-\frac 12-\frac 1{4n}$
For the general case
$$\sum\limits_{k=1}^n \left(\frac{1}{\left(2k-1+2^{1/\alpha}\right)^\alpha} - \frac{1}{\left(2k+2^{1/\alpha}\right)^\alpha}\right)=\sum\limits_{k=1}^n \left(\frac{\left(2k+2^{1/\alpha}\right)^\alpha-\left(2k-1+2^{1/\alpha}\right)^\alpha}{\left(2k-1+2^{1/\alpha}\right)^\alpha\left(2k+2^{1/\alpha}\right)^\alpha}\right)$$
If we are interested in how the sum grows asymptotically with $n$ we can assume the lower limit is rather large instead of $1$. Then the $2^{1/\alpha}$s will not matter and the numerator becomes $(2k)^\alpha(1-(1-\frac 1{(2k)})^\alpha)\approx \alpha(2k)^{\alpha-1}$.  With an upper limit of $\infty$ the sum will converge for all $\alpha \gt 0$.  To find the dependence at $\infty$ we will study the sum from $n$ to infinity instead of $1$ to $n$ by converting to an integral.
$$\sum\limits_{k=n}^\infty \left(\frac{\left(2k+2^{1/\alpha}\right)^\alpha-\left(2k-1+2^{1/\alpha}\right)^\alpha}{\left(2k-1+2^{1/\alpha}\right)^\alpha\left(2k+2^{1/\alpha}\right)^\alpha}\right)\approx \sum_n^\infty \frac {\alpha (2k)^{\alpha-1}}{(2k)^{2\alpha}}\\
=\sum_n^\infty \alpha (2k)^{-1-\alpha}\approx \int_n^\infty\alpha(2x)^{-1-\alpha}dx\\=\frac {n^{-\alpha}}{2^{1+\alpha}\alpha}$$ so the sum decreases from the limit at $\infty$ (whatever it is) like this.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{k = 1}^{\infty}{1 \over \pars{2k + s}^{\alpha}} & =
2^{-a}\sum_{k = 1}^{\infty}{1 \over \pars{k + s/2}^{\alpha}} =
2^{-a}\zeta\pars{a,{s + 2 \over 2}}\,,\quad\Re\pars{\alpha} > 1\,,\quad
s \not= 0,-2,-4,\ldots
\end{align}

where $\ds{\zeta}$ is the Hurwitz Zeta Function. Now, you can finish your exercise.

If you're still interested in the asymptotic behaviour, you can see
this identity.
