I am trying to work out what this series evaluates to: $$\sum_{i=1}^\infty i(1-k){k^{i-1}} $$ where k is a constant such that 0 < k < 1.

To figure this out I expand the brackets to get:

$$\sum_{i=1}^\infty (i{k^{i-1}} -ik^i)$$ which is equivalent to: $$ \lim_{N\to \infty}(\sum_{i=1}^N(i{k^{i-1}} -ik^i)) $$

I then try and write out each term trying to get them to cancel out but I get:

$$ \lim_{N\to \infty} (1 -k+2k-2k^2+3k^2-...+Nk^{N-1}-(N-1)k^{N-1}-Nk^N)$$ = $$ \lim_{N\to \infty}(\sum_{i=0}^{N-1} k^i)+\lim_{N\to \infty}(-Nk^N)$$

I think I want to use the fact that $$\left\lvert k \right\rvert<0$$ to show that the larger terms go to zero but I am unclear how to do this. Thanks for any help in advance.

  • $\begingroup$ Since you're using limits, I'm assuming you know calculus? I want to post an answer but I'm unsure of whether it will be understandable. $\endgroup$ – Edward Jiang Mar 28 '16 at 19:45
  • $\begingroup$ My suggestion is to write the summation $\sum ik^i$ as a double sum. $\endgroup$ – hardmath Mar 28 '16 at 19:46

Omitting the factor $1-k$, $$S:=\sum_{i=1}^\infty i{k^{i-1}}=\sum_{i=0}^\infty (i+1){k^i}=1+\sum_{i=1}^\infty (i+1){k^i}=1+kS+\sum_{i=1}^\infty{k^i},$$


$$(1-k)S=1+\frac k{1-k}=\frac1{1-k}.$$

Note that the geometric sum is obtained by the same trick,




If $|k|<1$ you can interchange sums and differentiation: $\sum_j j k^j = k\sum_j \frac{dk^j}{dk} = k \frac{d}{dk} \sum_{j=1}^{\infty} k^j$


This answer assumes some familiarity with Taylor series.

Note that the Taylor series for $1/(1-x)$ around $x=0$ is $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$ Within the radius of convergence of this Taylor series we can freely integrate and differentiate term by term. In particular $$ \frac{d}{dx} \left( \frac{1}{1-x}\right) = \sum_{n=1}^\infty n x^{n-1} $$ Again, within the radius of convergence, we can multiply by $(1-x)$ and we know $$ (1-x)\frac{d}{dx} \left( \frac{1}{1-x}\right) = \sum_{n=1}^\infty n x^{n-1}(1-x)$$ Can you continue from here?


Another rearrangement method (with no derivatives). If $|k|<1$, then these series converge absolutely so rearrangement is legitimate. $$ \sum_{i=1}^\infty i(1-k)k^{i-1} =\sum_{i=1}^\infty (i{k^{i-1}} -ik^i) = \sum_{i=1}^\infty i{k^{i-1}} - \sum_{i=1}^\infty ik^i \\ =\sum_{i=0}^\infty (i+1)k^i - \sum_{i=1}^\infty ik^i = 1+\sum_{i=1}^\infty (i+1)k^i - \sum_{i=1}^\infty ik^i \\ =1+\sum_{i=1}^\infty \big[(i+1)-i\big]k^i = 1+\sum_{i=1}^\infty k^i = \frac{1}{1-k} $$


The identity I think you need, or rather, the one you are looking for, is the following:

$\sum_{i=0}^{n-1} i a^i = \frac{a-na^n+(n-1)a^{n+1}}{(1-a)^2}$

Which can be found among the numerous, ever-helpful identities on this page. I am not certain as to the proof, yet.

But just using shifted sums and what we know we don't need to even use that identity:

$f(k)=\sum_{i=1}^{∞} i k^i$


$f(k)=\sum_{i=2}^{∞} ({i-1}) k^{i-1}$

$f(k)=\sum_{i=2}^{∞} ({ik^{i-1}-k^{i-1}})$

$\sum_{i=1}^{∞} i k^i = \sum_{i=2}^{∞} ({ik^{i-1})-\sum_{i=2}^{∞} (k^{i-1}})$

$\sum_{i=2}^{∞} (k^{i-1})=\sum_{i=2}^{∞} (ik^{i-1})-\sum_{i=2}^{∞} i k^i$

$1-k+\sum_{i=2}^{∞} (k^{i-1})=\sum_{i=1}^{∞} (ik^{i-1})-\sum_{i=1}^{∞} i k^i$

So, the original sum is found to be equal to:

$1-k+\sum_{i=1}^{∞} k^i$

The last summation being a geometric series with ratio $k$, whose sum to infinity is well known: $\frac{a}{1-r}$, where $a$ is the initial term, and $r$ is the ratio.

So my answer (though I might've made a mistake along the way) is $\frac{1-k+k^2}{1-k}$, converging for $-1<k<1$.

Let me know if anything conflicts with that I will amend the solution.



  • $$\sum_{n=a}^{m}nk^{n-c}\left(b-k\right)=\frac{(b-k)k^{-c}\left(k^a(a+k-ak\right)+k^{1+m}(m(k-1)-1)}{\left(k-1\right)^2}$$

  • And when $m\to\infty$ and $|k|<1$:


Now, when $a=b=c=1$:

  • $$\sum_{n=1}^{m}nk^{n-1}\left(1-k\right)=\frac{k^m(1+m-km)-1}{k-1}$$

Now, when $m\to\infty$ and $|k|<1$:

  • $$\sum_{n=1}^{\infty}nk^{n-1}\left(1-k\right)=\lim_{m\to\infty}\frac{k^m(1+m-km)-1}{k-1}=\frac{1}{1-k}$$
  • 1
    $\begingroup$ Seriously... do you think anybody can "notice" the first identity in your post? $\endgroup$ – Did Mar 28 '16 at 20:17

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