What would the infinite sum of c^n/n converge to?

Question

I know for example that the infinite sum of $c^n$ can be calculated when $|c|<1$ as below:

\begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} c^n=\frac{1}{1-c} \end{equation*}

And that the infinite sum of 1/n converges to infinity. But what is the sum of cⁿ/n?

\begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} \frac{c^n}{n} \end{equation*}

And would this depend on whether or not c is less than or greater than or equal to 1?

Answer 1

First note that your sum is only valid for $n \ge 1$ since you cannot divide by $0$.

Here is one approach to calculating that. Let $$ F(x) = \sum_{n=1}^\infty \frac{x^n}{n} $$ and differentiate it within the radius of convergence to get $$ F'(x) = \frac{d}{dx} \left[ \sum_{n=1}^\infty \frac{x^n}{n} \right] = \sum_{n=1}^\infty \frac{d}{dx} \left[ \frac{x^n}{n} \right] = \sum_{n=1}^\infty x^{n-1} $$ which you can sum as an ordinary geometric series and integrate back to get $F(x)$...

Answer 2

Your first series' answer is valid iff $|c| < 1$.

Similarly, in this case, if $|c| > 1$, then the series won't converge. In fact, $c^n/n \not\to 0$ as $n\to\infty$.

For $|c| < 1$, you have that $\displaystyle\sum_{n=1}^\infty \dfrac{c^n}{n}$ converges, by the comparison test.

For $c = 1$, the series does not converge. This is the harmonic series.

However, $c = -1$ does in fact converge. The sum is $-\log 2$. This is the alternating harmonic series.

Answer 3

You know that for $ c\in (-1,1),$

$$\sum_{n=0}^{+\infty}c^n=\frac{1}{1-c}$$

by integration over $[0,c] \subset (-1,1)$,

$$\sum_{n=0}^{+\infty}\frac{c^{n+1}}{n+1}=-\ln(1-c)$$ $$=\sum_{n=1}^{+\infty}\frac{c^n}{n}$$