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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.

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  • $\begingroup$ It's a differentiated geometric series, $\sum n\cdot x^{n-1}$. $\endgroup$
    – Daniel Fischer
    Commented Dec 7, 2014 at 16:57
  • $\begingroup$ See this for ideas. $\endgroup$
    – David Mitra
    Commented Dec 7, 2014 at 17:00
  • $\begingroup$ @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? $\endgroup$
    – Matt
    Commented Dec 7, 2014 at 17:03
  • $\begingroup$ 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? $\endgroup$
    – Daniel Fischer
    Commented Dec 7, 2014 at 17:07
  • $\begingroup$ 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? $\endgroup$
    – Matt
    Commented Dec 7, 2014 at 17:13

2 Answers 2

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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 $T$.

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  • $\begingroup$ It's amazing the many different ways one can solve some of these problems (I looked at the other links suggested in the comments to my question). How did you know to use the common ratio of 1/4 to "create" the geometric series? I'm guessing it's because of the common factor in each term of the original. I just never would thought to take a common factor and use it to create a simple geometric series like this. Very cool. $\endgroup$
    – Matt
    Commented Dec 7, 2014 at 18:06
  • $\begingroup$ After you understand this, try to do $$\sum\frac{n^2}{4^{n-1}}$$ $\endgroup$
    – GEdgar
    Commented Dec 7, 2014 at 20:40
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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 $$\sum_{n=1}^{\infty}a^{n-1}\sum_{i=1}^{\infty}a^{i-1}=\sum_{n=1}^{\infty}a^{n-1}\frac{1}{1-a}=\frac{1}{1-a}\sum_{n=1}^{\infty}a^{n-1}=(\frac{1}{1-a})^2$$

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