Sum of series $\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$ Find the radius of convergence and the sum of power series 
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
Radius of convergence is $R=1$, and the interval of convergence is $-1<x<1$.
I am having trouble in finding the sum.
Here is what I have tried.
$$\sum_{n=0}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\sum_{n=1}^{+\infty}(n+1)x^{2n+1}\int_0^1t^{2n}dt=\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)x^{2n+1}t^{2n}\right)dt$$
$$=x\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}\right)dt$$
$$\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}=\sum_{n=1}^{+\infty}n(xt)^{2n}+\sum_{n=1}^{+\infty}(xt)^{2n}$$
$$\sum_{n=1}^{+\infty}(xt)^{2n}=\frac{(xt)^2}{1-(xt)^2}$$
How to find the sum of $$\sum_{n=1}^{+\infty}n(xt)^{2n}?$$
EDIT:
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x^2\sum_{n=1}^{+\infty}\frac{x^{2n}}{2n+1}$$
$$=\frac{1}{2}x^2\sum_{n=1}^{+\infty}x^{2n}\int_0^1t^{2n}dt=\frac{1}{2}x^2\int_0^1\left(\sum_{n=1}^{+\infty}(xt)^{2n}\right)dt$$
$$=\frac{1}{2}x^2\int_0^1\frac{(xt)^2}{1-(xt)^2}=\frac{1}{2}x^4\int_0^1\frac{t^2}{1-(xt)^2}dt$$
$$=\frac{1}{2}x^4\cdot\frac{1}{x^3}(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|)\Rightarrow$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x\left(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|\right)\Rightarrow$$
$$\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)\Rightarrow$$
Finally,
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)$$
Question: Is this correct?
 A: Hint:
$$f(x)=\sum_{n=0}^\infty\frac{n+1}{2n+1}x^{2n+1}$$
$$\implies f'(x)=\sum_{n=0}^\infty(n+1)x^{2n}$$
And now use the geometric series
$$\frac1{1-r}=\sum_{n=0}^\infty r^n$$
$$\frac d{dr}\frac1{1-r}=\sum_{n=0}^\infty(n+1)r^n$$
A: 
How to find the sum of  $\displaystyle \,\sum_{n=1}^{+\infty}n(xt)^{2n}\,?$

One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1, \tag1
$$ then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ and by making $n \to +\infty$ in $(2)$, using $|x|<1$, multiplying by $x$, gives 

$$
\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2},\qquad |x|<1. \tag3
$$

A: $$S=\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
For the converge use Ratio Test.
$$2S=\sum_{n=1}^{+\infty}\frac{2n+1+1}{2n+1}x^{2n+1}=\sum_{n=1}^{+\infty}x^{2n+1}+\sum_{n=1}^{+\infty}\frac{x^{2n+1}}{2n+1}$$
The first sum is clearly an Infinite Geometric Series.
Now for the second part $$\ln(1+x)-\ln(1-x)=2\sum_{n=1}^{+\infty}\frac{x^{2n+1}}{2n+1}$$
A: Your work is good. Here's another way to do it.
Let,
$$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{2n+1}$$
Clearly we have,
$$f'(x)=\sum_{n=1}^{\infty} x^{2n}=\frac{1}{1-x^2}-1$$
It follows,
$$f(x)=\tanh^{-1}(x)-x$$
Now let,
$$g(x)=\sum_{n=1}^{\infty} n\frac{x^{2n+1}}{2n+1}$$
Clearly we have,
$$2g(x)+f(x)=\sum_{n=1}^{\infty} x^{2n+1}$$
$$2g(x)+\tanh^{-1}(x)-x=\frac{x}{1-x^2}-x$$
From which we get,
$$g(x)=\frac{x}{2-2x^2}-\frac{1}{2}\tanh^{-1}(x)$$
We are interested in $g(x)+f(x)$ this is,
$$\sum_{n=1}^{\infty}\frac{n+1}{2n+1}x^{2n+1}=\frac{1}{2}\tanh^{-1}(x)+\frac{x}{2-2x^2}-x$$
$$=\frac{1}{4}\ln (x+1)-\frac{1}{4}\ln (1-x)+\frac{x}{2-2x^2}-x$$
Edit
Note another way:
Assume $|x|<1$ and $|t|<1$.
$$\int_{0}^{x} t^{2n} dt=\frac{x^{2n+1}}{2n+1}$$
So that
$$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{2n+1}$$
$$=\int_{0}^{x} \sum_{n=1}^{\infty} t^{2n} dt$$
$$=\int_{0}^{x} \left(\frac{1}{1-t^2}-1 \right) dt $$
$$=\text{tanh}^{-1}(x)-x$$
It also follows that,
$$\frac{1}{2}x^2\sum_{n=1}^{+\infty}\frac{x^{2n}}{2n+1}$$
$$=\frac{1}{2}xf(x)$$
$$=\frac{1}{2}x\text{tanh}^{-1}(x)-\frac{1}{2}x^2$$
So this way works too. Just continue with your work or the previous solution.
