Find the positive values of $a$ such that the sequence: $$x_n=\left(1+\frac{1}{(n+1)^a}\right)\left(1+\frac{1}{(n+2)^a}\right)\ldots\left(1+\frac{1}{(2n)^a}\right)$$ is convergent.
I tried to make think about $\ln(x_n)$ but it didn't work and then I tried to write every memeber of the product to look like the Euler's sequence,but still couldn't find the values of $a$. Can somebody help me,please?
On
We have the for $a \le 0$ the sequence diverges, for $a>0$
$$\prod_{k=1}^n \left(1+\frac{1}{(n+k)^a}\right)=\sum_{k=1}^n \log\left(1+\frac{1}{(n+k)^a}\right)\sim \sum_{k=1}^n \frac{1}{(n+k)^a}=\frac1{n^a}\sum_{k=1}^n \frac{1}{(1+k/n)^a}$$
and since by Riemann sum
$$\lim_{n\to \infty}\frac1n\sum_{k=1}^n \frac{1}{(1+k/n)^a}=\int_0^1\frac1{(1+x)^a}dx=I<\infty$$
we have that $$\frac1{n^a}\sum_{k=1}^n \frac{1}{(1+k/n)^a}=\frac1{n^{a-1}}\frac1n\sum_{k=1}^n \frac{1}{(1+k/n)^a} \to \begin{cases}0\quad \text{for} \quad a>1\\I\quad \text{for} \quad a=1\\\infty\quad \text{for} \quad 0<a<1\end{cases}$$
The following inequality will be very useful: $$ (\log 2)\,x\le\log(1+x)\le x,\quad 0\le x\le 1. $$ From it we get $$ \frac{(\log 2)\,n}{(2\,n)^a}\le\log x_n\le\frac{n}{(n+1)^a}. $$ It follows that $$ \lim_{n\to\infty}x_n=\begin{cases}1 & \text{if }a>1,\\ \infty& \text{if }0<a<1.\end{cases} $$ The only remaining case is $a=1$. Then $x_n\le e$ and it is easy to see that $x_n$ is increasing, proving that it converges.