Studying this sum: $\sum_{k=0}^{\infty} (k+1)(x)^k$

Question

$$\sum_{k=0}^{\infty} (k+1)(x)^k\Longrightarrow x=-\frac{9}{10} \Longrightarrow\sum_{k=0}^{\infty} (k+1)\left(-\frac{9}{10}\right)^k=\frac{100}{361}\approx 0.2777$$ $$\sum_{k=0}^{\infty} (k+1)(x)^k\Longrightarrow x=-\frac{99}{100} \Longrightarrow\sum_{k=0}^{\infty} (k+1)\left(-\frac{99}{100}\right)^k=\frac{10000}{39601}\approx 0.2525$$ $$\sum_{k=0}^{\infty} (k+1)(x)^k\Longrightarrow x=-\frac{999}{1000} \Longrightarrow\sum_{k=0}^{\infty} (k+1)\left(-\frac{999}{1000}\right)^k=\frac{1000000}{3996001}\approx 0.25025$$ $$\sum_{k=0}^{\infty} (k+1)(x)^k\Longrightarrow x=-\frac{99999999}{10^{8}} \Longrightarrow\sum_{k=0}^{\infty} (k+1)\left(-\frac{99999999}{10^{8}}\right)^k=\frac{10^{16}}{39999999600000001}\approx 0.2500000025$$

So my conclusion will be that when $x$ get to $-1$ the sum becomes $\frac{1}{4}$ so than this will be equal:

$$\lim_{x\rightarrow-1}\left(\sum_{k=0}^{\infty} (k+1)(x)^k\right)=\lim_{x\rightarrow-1}\left(\frac{1}{(1-x)^2}\right)=\frac{1}{\lim_{x\rightarrow-1} (1-x)^2}=\frac{1}{4}$$

and that is than equal to:

$$1-2+3-4+5-6+7-8+9-10+...=\frac{1}{4}$$

Can someone tell me if this is right? And if it is not can you proof it?

Answer 1

The terms of the form $(k+1)x^k$ are convergent to zero for $x$ in the interval $(-1,1)$. So the limit for their sum, as you've calculated, is correct.

That being said, you cannot then deduce the last summation, as you're applying it to a point where the terms of the sum are not convergent. The expression only makes sense on the open interval, and so cannot be be evaluated at the boundary.

Answer 2

You're looking at $\sum\limits_{k=0}^\infty (k+1)x^k$. One easily identifies this as the taylor series of $$f(x)=\frac{1}{(x-1)^2}$$ at $x=0$. Although the domain of this function is $\mathbb R\setminus\{1\}$, the taylor series does not need to converge on the domain but only on the radius of convergence. In this case we get $r=1$, meaning that we can only say that the taylor series converges for $|x|<1$ and for these $x$ the taylor series converges to the value of $f$.

For $|x|=1$ we can't make any statement yet and you have to manually check for convergence. In this case you get $\sum\limits_{k=0}^\infty (k+1)$ and $\sum\limits_{k=0}^\infty (k+1)\cdot(-1)^k$ and these series do not converge. This means, that you can't extend the taylor series to represent the function on $[-1,1]$ and therefore you do not get $$1-2+3-\dots=\frac 14.$$