# 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.

• It's a differentiated geometric series, $\sum n\cdot x^{n-1}$. Commented Dec 7, 2014 at 16:57
• See this for ideas. Commented Dec 7, 2014 at 17:00
• @DanielFischer, Ah is see that now as you point that out. We didn't talk about this possibly in class. It would be easy to handle $\sum [\frac{1}{4}]^{n} = \frac{4}{3}$. I also see that if I square this I get $\frac{16}{9}$. Is this insight by coincidence?
– Matt
Commented Dec 7, 2014 at 17:03
• No, you know that $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$ for $\lvert x\rvert < 1$. If you differentiate both sides, what do you get? Commented Dec 7, 2014 at 17:07
• I get: $nx^{n-1} = -\frac{1}{(1-x)^2}$. So I see the relationship thanks as this works out to $16/9$ when x is $1/4$. I guess the negative sign doesn't matter?
– Matt
Commented Dec 7, 2014 at 17:13

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 $T$.
• After you understand this, try to do $$\sum\frac{n^2}{4^{n-1}}$$ Commented Dec 7, 2014 at 20:40
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 $$\sum_{n=1}^{\infty}a^{n-1}\sum_{i=1}^{\infty}a^{i-1}=\sum_{n=1}^{\infty}a^{n-1}\frac{1}{1-a}=\frac{1}{1-a}\sum_{n=1}^{\infty}a^{n-1}=(\frac{1}{1-a})^2$$