I've been working on understanding series and their summations, and I came across this particular series: $$\alpha+2\alpha^2+3\alpha^3+...$$

The solution provided states that the sum of this series is $$\frac{\alpha}{(\alpha-1)^2}$$, but I'm having trouble understanding why this is the case.

My Work:

I tried to approach this problem by comparing it to the geometric series summation formula, which is $$\frac{a}{1-r}$$ where 'a' is the first term and 'r' is the common ratio. However, I'm not sure how to apply it here since the coefficients of $$\alpha$$ are increasing.


I'm currently an undergraduate student taking a course in Calculus II. We've covered sequences and series, including geometric series and the tests for convergence.


Here, $$\alpha$$ is a real number and the series is presumably infinite.

I would appreciate any help or guidance on how to approach this problem. Thank you in advance!

  • 3
    $\begingroup$ It's $\alpha \dfrac d{d\alpha}\dfrac1{1-\alpha}$ $\endgroup$ May 15 at 15:16
  • 1
    $\begingroup$ To add to @J.W.Tanner's comment, you'll also need to know about term-by-term differentiation of series. $\endgroup$
    – Umberto P.
    May 15 at 15:19

3 Answers 3


It's called an arithmetico-geometric series and can be reduced to a more familiar geometric series as follows:

\begin{align} S \,&=\, \alpha + 2\alpha^2 + 3\alpha^3 + \cdots\\ \alpha S \,&=\, \phantom{\alpha + 2}\alpha^2+2\alpha^3 + \cdots\\ S - \alpha S \,&=\, \alpha + \phantom{2}\alpha^2 + \phantom{3}\alpha^3 +\cdots\\ \end{align}

This last line is a geometric series, so we have \begin{align} S\left(1-\alpha\right) \,&=\, \displaystyle\frac{\alpha}{1-\alpha}\\\\ S \, &=\, \displaystyle\frac{\alpha}{\left(1-\alpha\right)^2} \end{align}

Remember that this is only true when the series converges, which occurs for $|\alpha| < 1$.


For finite and infinite sums:

This series can be solved quite simply: $$ f(\alpha)=\sum_{n=1}^kn\alpha^{n-1}\implies \int f(\alpha) {d\alpha}=\sum_{n=1}^k \alpha^{n}=\frac{(1-\alpha^k)}{1-\alpha}\implies f(\alpha) = -\frac{k\alpha^{k-1}}{1-\alpha}+\frac {1-\alpha^k}{(1-\alpha)^2} $$ But our actual sum is: $$ g(\alpha)=\alpha+2\alpha^2 +\dots +k\alpha^k= \sum_{n=1}^k n\alpha^n $$ But $$f(\alpha)=\sum_{n=0}^k (n+1)\alpha^n = g(\alpha)+\frac{1-\alpha^k}{1-\alpha}-1\implies g(\alpha)=1+\frac{1-\alpha^k}{(1-\alpha)^2}-\frac{1-\alpha^k}{(1-\alpha)}-\frac{k\alpha^{k-1}}{1-\alpha}=\frac{\alpha^{k+1}(\alpha k - k -1) + \alpha}{(\alpha-1)^2} $$


For a "direct" explanation that does not rely on differentiation, consider multiplying $(\alpha-1)^2$ by $\alpha+2\alpha^2+3\alpha^3+\cdots$.


(This is of course assuming the infinite series on the right converges in the first place.)

Now in the product, for example what would the coefficient of $\alpha^{42}$ be? There would be contribution from $\alpha^2\cdot40\alpha^{40}$, $-2\alpha\cdot41\alpha^{41}$, and $1\cdot42\alpha^{42}$. That all adds up to $0$.

There is nothing special about $42$. The same thing happens for all of the terms in the product, starting with the degree $3$ term. But we will directly examine the $\alpha$ and $\alpha^2$ terms in the product since those powers are not high enough to have the same pattern apply.

For $\alpha$, there is only the $1\cdot\alpha$ contribution. So there is an $\alpha$ term in the product.

For $\alpha^2$, there are contributions from the $-2\alpha\cdot\alpha$ and $1\cdot2\alpha^2$. Those cancel out.

So we have: $$\left(\alpha^2-2\alpha+1\right)\left(\alpha+2\alpha^2+3\alpha^3+\cdots\right)=\alpha$$ from which you get the result you were investigating. Again, it's only valid when $\alpha+2\alpha^2+3\alpha^3+\cdots$ converges in the first place though, which happens to be for $\alpha\in(-1,1)$.