# Thread: Exchangeable of the limitation

1. ## Exchangeable of the limitation

$\lim_{x\rightarrow1-0}\sum_{n=1}^{\infty}\frac{x^{n-1}}{nln(n+1)} = \sum_{n=1}^{\infty}\lim_{x\rightarrow1-0}\frac{x^{n-1}}{nln(n+1)}$

Is the above equation right? in the other words, can we take the limitation into the summation?

2. Yes, as long as the quantity considered is positive. The worse thing that could happen is that both iterated limits are infinite.

3. Originally Posted by Xingyuan
$\lim_{x\rightarrow1-0}\sum_{n=1}^{\infty}\frac{x^{n-1}}{nln(n+1)} = \sum_{n=1}^{\infty}\lim_{x\rightarrow1-0}\frac{x^{n-1}}{nln(n+1)}$

Is the above equation right? in the other words, can we take the limitation into the summation?
This would be true if $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n\log(n+1)}$ converged. As it is stated it doesn't make much sense. But, for example $\displaystyle \lim_{x\to 1^{-1}}\sum_{n=1}^{\infty}\frac{x^{n-1}}{n^2\log(n+1)}=\sum_{n=1}^{\infty}\lim_{x\to 1^{-}}\frac{x^{n-1}}{n^2\log(n+1)}$. This is because $\displaystyle \frac{x^{n-1}}{n^2\log(n+1)}$ has radius of convergence $1$ from where you can apply Abel's Limit Theorem.