Is it possible to find the sum of the infinite series $1/p + 2/p^2 + 3/p^3 + \cdots + n/(p^n)+\cdots$, where $p>1$? Is it possible to find the sum of the series:
$$\frac{1}{p} + \frac{2}{p^2} +\frac{3}{p^3} +\dots+\frac{n}{p^n}\dots$$ 
Does this series converge? ($p$ is finite number greater than $1$)
 A: Hint:
$$\frac{1}{1-x}=1+x+x^2+\dots$$
Differentiate this formula (it is uniformly convergent for $|x|<1$) and multiply with x. Finally set $x=1/p$.
A: The exact value is $\frac{\frac{1}{p}}{(1-\frac{1}{p})^{2}}$ when $-1<\frac{1}{p}<1$. Since $p>1$ in this case, the series converges.
Proof:
$$
\frac{1}{p}+\frac{2}{p^{2}}+\frac{3}{p^{3}}...
$$
$$
=(\frac{1}{p}+\frac{1}{p^{2}}+\frac{1}{p^{3}}...)+(\frac{1}{p^{2}}+\frac{1}{p^{3}}+\frac{1}{p^{4}}...)+(\frac{1}{p^{3}}+\frac{1}{p^{4}}+\frac{1}{p^{5}}...)...
$$
$$
=\frac{\frac{1}{p}}{1-\frac{1}{p}}+\frac{\frac{1}{p^{2}}}{1-\frac{1}{p}}+\frac{\frac{1}{p^{3}}}{1-\frac{1}{p}}...
$$
$$
=\frac{\frac{\frac{1}{p}}{1-\frac{1}{p}}}{1-\frac{1}{p}}
$$
$$
=\frac{\frac{1}{p}}{(1-\frac{1}{p})^{2}}
$$
A: For $p>1$ the series converges.
Lets take the general geometric series;
$$|x|<1\\a_1(1+x+x^2+x^3+...)=\frac{a_1}{1-x}\\a_1(1+2x+3x^2+4x^3+...)=a_1(1+x+x^2+x^3+...)'=(\frac{a_1}{1-x})'=\frac{a_1}{(1-x)^2}$$
So in our case;
$$a_1=\frac{1}{p}\ \ ,\ \ x=\frac{1}{p}\\\frac{1}{p}+\frac{2}{p^2}+...=\frac{1}{p}\cdot\frac{1}{(1-\frac{1}{p})^2}=\frac{1}{p(1-\frac{2}{p}+\frac{1}{p^2})}=\frac{1}{\frac{p^2-2p+1}{p}}=\frac{p}{(p-1)^2}$$
A: Hint: Consider the function $f(x) = x+x^2+x^3+....$, $|x| < 1$. We have $xf'(x) = x + 2x^2+3x^3+....$, and $f(x) = \dfrac{x}{1-x}\implies f'(x) = ...$. Thus you can find the sum of the right hand side, and all you need is to plug $x = \dfrac{1}{p}$ into the formula to get the answer.
A: Let:
$
S = {1\over p} + {2\over p^2} + {3 \over p^3} + \cdots \\
\implies {S\over p} = {1\over p^2} + {2\over p^3} + {3\over p^4} + \cdots.
$
Hence,
$
S\left ( 1 - {1\over p} \right ) = {1\over p} + {1\over p^2} + {1\over p^3} + \cdots \\
S\left ( 1 - {1\over p} \right ) = \frac{1}{p}\frac{1}{1-{1\over p}} = \frac{1}{p-1}\\
\implies S = \frac{p}{(p-1)^2}.
$
The series converges for $p>1$.
