Evaluate the following sums:

  1. $\sum\limits_{i=0}^\infty\frac1{4^i}$.
  2. $\sum\limits_{i=0}^\infty\frac i{4^i}$.
  3. $\sum\limits_{i=0}^\infty\frac {i^2}{4^i}$.
  4. $\sum\limits_{i=0}^\infty\frac {i^N}{4^i}$.

I have to do the problems above without using derivatives. I have done the first one and second one. I am stuck on the third one. I don't have to do the last one. Any help as to how to tackle this problem without differentiating (everyone who has helped me has used calc but we aren't allowed to do that!)

I did the first one using 1/(1-A) and got 7/6.

For the second one, I wrote out the series, S. Then i multiple all of the series by 7 and got 7S. I then subtracted S from 7S and was left with a series I could use 1/(1-A) on and found that the answer was 7/36.

How would I use a similar method to obtain #3?


  • 1
    $\begingroup$ Use your method again, just more of it. Take the series $S$, multiply by $4$ to get $4S$, then multiply by $4$ again to get $4^2S$. Now combine these three somehow. $\endgroup$ – GEdgar Sep 15 '14 at 21:18
  • 1
    $\begingroup$ See here. $\endgroup$ – Mhenni Benghorbal Sep 15 '14 at 21:51

To make manipulations more tractable, I'll consider $S_N(x)=\sum_{i=0}^\infty i^N x^i$ for $N=0,1,2$ and leave the substitution of $x=1/4$ to the reader. For $N=0$ we have the usual geometric sum $S_0(x)=\dfrac{1}{1-x}$. The familiar proof of this can be boiled down to observing that

\begin{align} (1-x)S_0(x)=S_0(x)-x S_0(x) =1&+x+x^2+\cdots\\ &-x-x^2-\cdots=1. \end{align}

This contains an insight which generalizes to any given power series $A(x)=\sum_{i=0}^\infty a_i x^i$:

\begin{align} (1-x)A(x)=A(x)-x A(x) =a_0&+a_1 x + a_2 x^2+\cdots\\ &-a_0x-a_1x^2-\cdots=a_0+\sum_{i=1}^\infty (a_{i+1}-a_i)x^i \end{align} In other words, multiplying by $(1-x)$ replaces the coefficients of $A(x)$ with the first differences of these coefficients (which are all zero for the geometric series save the zeroth.) Multiplying by further factors of $(1-x)$ gives the second differences, third differences, and so forth.

For the case of $N=1$, we note that the sequence $\{0,1,2,3,\ldots\}$ has first differences $\{0,1,1,1,\cdots\}$ and second differences $\{0,1,0,0,\ldots\}$. Consequently $S_1(x)=\dfrac{x}{(1-x)^2}$. For $N=2$, we need to go up to third differences:

$$\{0,1,4,9,16,\ldots\}\to \{0,1,3,5,7,\ldots\}\to \{0,1,2,2,2,\ldots\}\to \{0,1,1,0,0,\ldots\}$$

and from this we can read off the form of $S_2(x)$ directly (what is it?). The obvious pattern here is that the $(N+1)$-th differences of $\{i^n\}$ have finitely many nonzero entries, and so $(1-x)^{N+1}S_N(x)$ is some polynomial in $x$. (Actually finding this polynomial isn't so nice since it requires computing up to the $(N+1)$ difference, so this really only works well if $a_n$ is a polynomial of low degree in $n$.)


The first one is a geometric series: $~\displaystyle\sum_{k=0}^\infty x^k=\frac1{1-x},~$ where $x=\dfrac14.~$ Now, what would happen

if you were to differentiate-and-then-multiply both sides with regard to x once, twice, thrice, etc. ?


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