Is this a geometric series? If so please help point me in the right direction for calculating the sum: $\sum_{n=1}^\infty n[{1\over 4}]^{n-1}$ Is this a geometric series? If so please help point me in the right direction for calculating the sum: $$\sum_{n=1}^\infty n[{1\over 4}]^{n-1}$$
I know using the test for divergence that this does not diverge. Also using the root (or ratio) test it will converge. I can see using a table that with enough terms the sum appears to be going towards $16/9$.
How though do I solve this problem using calculus? I can't use the simple geometric formula $s = \frac{a}{1-r}$ because there is no common ratio (the ratio for each term goes from $\frac{2}{4}$, to $\frac{3}{8}$, $\frac{4}{12}$, $\frac{5}{16}$, $\frac{6}{20}$, ...).
So I ask is this a geometric series (the problem I'm working on says it is and I want to double check). If so, then how do I figure out the formula representing the partial sums so that I can later take the limit of the sequence of partial sums to find the answer.
 A: An answer for a less advanced reader.
Suppose we know it converges.  Say it converges to $T$.  Then
$$
T = \frac{1}{1}+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+\frac{5}{4^4}+\dots
$$
Divide by $4$
$$
\frac{T}{4} = \frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+\frac{5}{4^5}+\dots
$$
subtract... for each denominator, subtract the two terms with that denominator:
$$
T - \frac{T}{4} =
1 +\frac{2-1}{4}+\frac{3-2}{4^2}+\frac{4-3}{4^3}+\frac{5-4}{4^4}+\dots
\\
=1 +\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\dots
$$
Now this is a geometric series.  So evaluate it as $4/3$
to get
$$
\frac{3}{4}\;T = \frac{4}{3}
$$
which you can solve for $T$.
A: Let $a=\frac 1 4$, now notice that $na^{n-1}= a^{n-1}+a^{n-1}+a^{n-1}...+a^{n-1}$ $n$ times,
so our series looks like 
$$\sum_{n=1}^{\infty} na^{n-1}=a^0+a^1+a^1+a^2+a^2+a^2...=a^0+a^1+a^2....+a^1+a^2+a^3...+a^2+a^3+a^4... ...=\sum_{n=1}^{\infty}\sum_{i=n}^{\infty}a^{i-1}=\sum_{n=1}^{\infty}a^{n-1}\sum_{i=1}^{\infty}a^{i-1}$$
Now $\sum_{i=1}^{\infty}a^{i-1}$ can be evaluated as 
$$\sum_{i=1}^{\infty}a^{i-1}=\frac{1}{1-a}$$
since $a<1$, thus
$$\sum_{n=1}^{\infty}a^{n-1}\sum_{i=1}^{\infty}a^{i-1}=\sum_{n=1}^{\infty}a^{n-1}\frac{1}{1-a}=\frac{1}{1-a}\sum_{n=1}^{\infty}a^{n-1}=(\frac{1}{1-a})^2$$
