# Understanding the Series Summation: $\alpha+2\alpha^2+3\alpha^3+...=\frac{\alpha}{(\alpha-1)^2}$ [duplicate]

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.

Background:

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.

Definitions:

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!

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

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 $$\alpha+2\alpha^2+3\alpha^3+\cdots$$.

$$\left(\alpha^2-2\alpha+1\right)\left(\alpha+2\alpha^2+3\alpha^3+\cdots\right)$$

(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 $$\alpha\in(-1,1)$$.