Check if convergent in the end points of the convergence interval I have to find the radius of convergence for $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}x^n$, for x$\in$R.
I think it is $r^{-1}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{ \frac{n+1}{n+2} }{ \frac{n}{n+1}}=1$ so r=1.
Then I have to find out if $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}x^n$ is convergent in 
the end points of the convergence interval. I think I have to check for $|r|$.
But we can't use ratio test (L=1) or integral test(not decreasing) for $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}$. How can I then check if it's convergent in the end points of the convergence interval?
 A: You found that the radius of convergence is 1. That is true, and now we need to check at the edges, i.e let $x=\pm 1$.
$$\sum_{n=0}^{\infty}\frac{n}{n+1}1^n=\sum_{n=0}^{\infty}\frac{n}{n+1}$$ This diverges because $\lim\frac{n}{n+1}=1\neq0$.
$$\sum_{n=0}^{\infty}\frac{n}{n+1}(-1)^n$$
This also diverges because $(-1)^n\frac{n}{n+1}$ diverges.
So we have the the series converges at $(-1,1)$
A: Indeed the radius of convergence is $r=1$. To figure out if the endpoints $-1$ and $1$ are part of the interval of convergence simply plug them in for $x$ in your series. For $x = 1$, we obtain the series $\sum_{n=0}^\infty \frac{n}{n+1}$, which evidently diverges by the Divergence Test. For $x=-1$, you have $\sum_{n=0}^\infty \frac{n}{n+1}(-1)^n$. Again, the Divergence Test tells us this diverges (the sequence $\{(-1)^n \frac{n}{n+1}\}$ doesn't converge at all). The interval of convergence is thus (-1,1).
A: ...or, you can simplify $\frac{n}{n+1} = 1 - \frac{1}{n+1}$  to get (1) the famous Geometric series, and then it converges for $|x|<1$, and (2) $\sum_k \frac{x^k}{k+1} = \frac{1}{x} \int_{0}^{x}[\sum_{k=0}^{\infty} y^{k} ]dx$. This, after a bit of manipulation becomes $-\frac{\log (1-x)}{x}$ which converges for $x<1$, so combined with the first sum the interval of convergence is $(-1,1)$. It might seem that $x=0$ is a problem, but in fact if you compute the limit 
$$
-\lim_{x \to 0} \frac{\log(1-x)}{x} = 1
$$
so the limit exists, so the function is continuous, and the interval holds.
