Sum of a series to an exact answer 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.
 A: Omitting the factor $1-k$,
$$S:=\sum_{i=1}^\infty i{k^{i-1}}=\sum_{i=0}^\infty (i+1){k^i}=1+\sum_{i=1}^\infty (i+1){k^i}=1+kS+\sum_{i=1}^\infty{k^i},$$
and
$$(1-k)S=1+\frac k{1-k}=\frac1{1-k}.$$

Note that the geometric sum is obtained by the same trick,
$$T:=\sum_{i=1}^\infty{k^i}=\sum_{i=0}^\infty{k^{i+1}}=k+\sum_{i=1}^\infty{k^{i+1}}=k+kT,$$
$$(1-k)T=k.$$
A: If $|k|<1$ you can interchange sums and differentiation: $\sum_j j k^j = k\sum_j \frac{dk^j}{dk} = k \frac{d}{dk} \sum_{j=1}^{\infty} k^j$
A: 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?
A: 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: 
$f(k)=\sum_{i=1}^{∞} i k^i$
Then 
$f(k)=\sum_{i=2}^{∞} ({i-1}) k^{i-1}$
$f(k)=\sum_{i=2}^{∞} ({ik^{i-1}-k^{i-1}})$
$\sum_{i=1}^{∞} i k^i = \sum_{i=2}^{∞} ({ik^{i-1})-\sum_{i=2}^{∞} (k^{i-1}})$
$\sum_{i=2}^{∞} (k^{i-1})=\sum_{i=2}^{∞} (ik^{i-1})-\sum_{i=2}^{∞} i k^i$
$1-k+\sum_{i=2}^{∞} (k^{i-1})=\sum_{i=1}^{∞} (ik^{i-1})-\sum_{i=1}^{∞} i k^i$
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.
A: Another rearrangement method (with no derivatives).  If $|k|<1$, then these series converge absolutely so rearrangement is legitimate.
$$
\sum_{i=1}^\infty i(1-k)k^{i-1}
=\sum_{i=1}^\infty (i{k^{i-1}} -ik^i) =
\sum_{i=1}^\infty i{k^{i-1}} - \sum_{i=1}^\infty ik^i
\\
=\sum_{i=0}^\infty (i+1)k^i - \sum_{i=1}^\infty ik^i =
1+\sum_{i=1}^\infty (i+1)k^i - \sum_{i=1}^\infty ik^i
\\
=1+\sum_{i=1}^\infty \big[(i+1)-i\big]k^i =
1+\sum_{i=1}^\infty k^i = \frac{1}{1-k}
$$
