How to prove $\lim \limits_{\alpha\rightarrow \infty} (\sum_{n=1}^{\infty}[(a_{n}+\frac{1}{(1+n)^{\alpha}})^{n}-{a}_{n}^{n}]) = 0 $

Question

${a_{n}}$ is a sequence satisfy $a_{n} >0$ and $\sum_{n=1}^{\infty} a_{n}<\infty$ ,$\alpha >0$，prove $\lim \limits_{\alpha\rightarrow \infty} (\sum_{n=1}^{\infty}[(a_{n}+\frac{1}{(1+n)^{\alpha}})^{n}-{a}_{n}^{n}]) = 0 $

I have got $\sum_{n=1}^{\infty}(a_{n}+\frac{1}{(1+n)^{\alpha}})^{n}$ and$\sum_{n=1}^{\infty}{a}_{n}^{n} $ both converge.But I do not know how to do next.Any suggestion is appreciated.

Answer 1

Note that

$$\left(a_{n}+\frac{1}{(1+n)^{\alpha}}\right) ^{n}=e^{n\log \left(a_{n}+\frac{1}{(1+n)^{\alpha}}\right)}$$

and

$$\log \left(a_{n}+\frac{1}{(1+n)^{\alpha}}\right)=\log a_n+\log \left(1+\frac{1}{a_n(1+n)^{\alpha}}\right)\sim\log a_n+\frac{1}{a_n(1+n)^{\alpha}} $$

thus

$$\left(a_{n}+\frac{1}{(1+n)^{\alpha}}\right) ^{n}=e^{n\log \left(a_{n}+\frac{1}{(1+n)^{\alpha}}\right)}\sim a_n^n\cdot e^{\frac{n}{a_n(1+n)^{\alpha}}}$$

and for $\alpha>1$

$$e^{\frac{n}{a_n(1+n)^{\alpha}}}\sim 1+\frac{n}{a_n(1+n)^{\alpha}}$$

therefore

$$\sum_{n=1}^{\infty}[(a_{n}+\frac{1}{(1+n)^{\alpha}})^{n}-{a}_{n}^{n}]\sim\sum_{n=1}^{\infty} \frac{na_n^{n-1}}{(1+n)^{\alpha}}$$