Sum of $1-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+....$ $1-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+....$ 
How can we find sum of above series upto infinite terms?
I don't know how to start and just need some hint.
 A: Say sum to $n$ terms of the series is $S_n$.
Now we have that $$S_n=1+\sum_{r=1}^n (-1)^n\cdot\frac{(n+1)^2}{5^{n}}=1+R_n$$
where $R_n=\sum_\limits{r=1}^n (-1)^n\cdot\frac{(n+1)^2}{5^{n}}=-\frac{2^2}{5}+\sum_\limits{r=1}^n (-1)^{n+1}\cdot\frac{(n+2)^2}{5^{n+1}}$
Now $$-\frac{1}{5}\cdot R_n=\sum_{r=1}^n (-1)^{n+1}\cdot\frac{(n+1)^2}{5^{n+1}}$$
Subtracting, we get $$\frac{6}{5}\cdot R_n=-\frac{2^2}{5}+\sum_{r=1}^n (-1)^{n+1}\cdot\frac{2n+3}{5^{n+1}}=-\frac{2^2}{5}+Q_n$$
Similarly $$Q_n=\frac{1}{5}+\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2n+5}{5^{n+2}}$$
and $$-\frac{1}{5}\cdot Q_n=\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2n+3}{5^{n+2}}$$
Again we get that $$\frac{6}{5}Q_n=\frac{1}{5}+\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2}{5^{n+2}}$$
When $n\to\infty$, we have that $$Q_\infty=\frac{5}{6}(\frac{1}{5}-\frac{1}{75})=\frac{7}{45}$$
Hence we can calculate $R_\infty$ and $S_\infty$ which comes out to be $\frac{25}{54}$.
A: Note that
$$\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \dots$$
$$\frac{d}{dx}\left(\frac{1}{1 - x}\right) = 1 + 2x + 3x^2 + 4x^3 + \dots$$
$$x\cdot\frac{d}{dx}\left(\frac{1}{1 - x}\right) =  x + 2x^2 + 3x^3 + 4x^4 + \dots$$
$$\frac{d}{dx}\left(x\cdot\frac{d}{dx}\left(\frac{1}{1 - x}\right)\right) = 1 + 2^2x + 3^2x^2 + 4^2x^3 + \dots$$
Now determine LHS and substitute in $x = -\frac{1}{5}$ (I used Wolfram): 
$$-\frac{x + 1}{(x - 1)^3} = 1 + 2^2x + 3^2x^2 + 4^2x^3 + \dots$$
$$\frac{25}{54} = 1 - \frac{2^2}{5} + \frac{3^2}{5^2} - \frac{4^2}{5^3} + \dots$$
