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

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

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- Oct 11th 2009, 07:17 PMXingyuanExchangeable of the limitation
$\displaystyle \lim_{x\rightarrow1-0}\sum_{n=1}^{\infty}\frac{x^{n-1}}{nln(n+1)} = \sum_{n=1}^{\infty}\lim_{x\rightarrow1-0}\frac{x^{n-1}}{nln(n+1)}$

Is the above equation right? in the other words, can we take the limitation into the summation? - Dec 30th 2010, 04:25 AMRebesques
Yes, as long as the quantity considered is positive. The worse thing that could happen is that both iterated limits are infinite.

- Dec 30th 2010, 09:21 AMDrexel28
This would be true if $\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{1}{n\log(n+1)}$ converged. As it is stated it doesn't make much sense. But, for example $\displaystyle \displaystyle \lim_{x\to 1^{-1}}\sum_{n=1}^{\infty}\frac{x^{n-1}}{n^2\log(n+1)}=\sum_{n=1}^{\infty}\lim_{x\to 1^{-}}\frac{x^{n-1}}{n^2\log(n+1)}$. This is because $\displaystyle \displaystyle \frac{x^{n-1}}{n^2\log(n+1)}$ has radius of convergence $\displaystyle 1$ from where you can apply Abel's Limit Theorem.