A Geometric or Harmonic Series?

Consider the geometric series (which holds for $-1<x<1$ ): $\frac{1}{1-x}=1+x+x^2+x^3+\ldots$

Integrate both sides: $C-\ln{(1-x)}=\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots$

Substitute $x=0$ to find that $C=0$ .

Thus, $\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots=-\ln{(1-x)}$

By the way, the sum does not diverge. This can be seen with $\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots<1+x+x^2+x^3+\ldots=\frac{1}{1-x}$ for $0≤x<1$ . In addition, in the cases $-1≤x<0$ , we can prove that it must converge using the Alternative Series Test.

Bonus

What is the general sum of $x+2x^2+3x^3+4x^4+\ldots$ ?

Original problem: (quick solution as it is the same method used by Nick). Let's suppose that $f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ . Then $f'(x) = \sum_{n=1}^{\infty} x^{n-1} = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ . So $f(x) = \int \frac{1}{1-x} = - \log (1-x) + C.$ Using the original definition of $f$ , if we evaluate the last expression for $x=0$ we get $0 = -0 + C \implies C=0$ so $f(x) = - \log(1-x)$ .

Note: To be completely rigorous, before the first step one should prove that defining $f$ in such a way is valid. To do this, we should prove the series converges. For $x \in (-1,1)$ the ratio test will do this. For $x=-1$ the alternating series test will do the trick.

Bonus: (Left by Nick in his solution) First, let's apply the ratio test to $\sum_{n=1}^{\infty} nx^{n}$ . We need to compute $\lim_{n \to \infty} |\frac{(n+1)x^{n+1}}{nx^n}| = \lim_{n \to \infty} |\frac{n+1}{n} x| \to |x|$ which is smaller than one precisely when $|x| < 1$ so it converges for $x \in (-1,1)$ . And for $|x| > 1$ it diverges. For |x|=1 divergence is also obvious, but a different method would need to be used. So we may define $f(x) = \sum_{n=1}^{\infty} nx^{n}$ as a function in $(-1,1)$ . We know that $f(0) = 0$ . For $x \neq 0$ we have $\frac{f(x)}{x} = \sum_{n=1}^{\infty} nx^{n-1}$ and thus $\int \frac{f(x)}{x} dx = \sum_{n=1}^{\infty} x^n = \frac{x}{1-x}$ . If we differentiate both sides we get $\frac{f(x)}{x} = \frac{1}{(1-x)^2}$ so $f(x) = \frac{x}{(1-x)^2}$ . Luckily for us, this new definition for $f$ agrees with the previously computed value of $f(0)$ .