$x_n = \sum_{k=1}^n ka^k$. Convergence and limit.

Question

Let $a\in=(-1;1)$ and $$x_n = \sum_{k=1}^n ka^k, n\ge1$$ Prove that $(x_n)_n$ is convergent and that has the limit equal to $\dfrac{a}{(1-a)^2}$

I was able to find the limit of the sum. For that I took the following function $$f:(-1;1)\to\mathbb{R}, f(a)=(\sum_{k=0}^{\infty} a^k) - 1=\dfrac{1}{1-a}-1$$ Taking derivatives of both forms we have: $$f'(a)=\dfrac{1}{(1-a)^2}$$ $$f'(a)=\sum_{k=0}^{\infty} ka^{k-1}$$ Multplying the second one by $a$ and having $f'(a)$ being subtitute we get the limit of the initial sum $\dfrac{a}{(1-a)^2}$

How can I show that $x_n$ is convergent? Computing the limit did I show that? Becuase the limit is finite. I want a rigorous proof. Thank you!

Answer 1

To show that $x_n$ is convergent, use the ratio test. $\lim_{k\to\infty}\frac{|ka^k|}{|(k-1)a^{k-1}|}=|a|<1$, so the series converges.

Answer 2

Consider the finite geometric series:

$$\sum_{k=0}^na^k=\frac{1-a^{n+1}}{1-a}$$

The derivative is given by

$$\sum_{k=0}^nka^{k-1}=\frac d{da}\frac{1-a^{n+1}}{1-a}$$

At $k=0$, we remove the first term. Multiply both sides by $a$ to get,

$$\sum_{k=1}^nka^k=a\left[\frac d{da}\frac{1-a^{n+1}}{1-a}\right]$$

And thus you have the explicit form of $x_n$. Can you show that for $|a|<1$, $x_n$ converges?