Convergence of this alternating series: $\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)C^k} = C \log \frac{C+1}{C}$

Question

I "heard" the following formula for any $C \ge 1$:

$\sum\limits_{k=0}^\infty \dfrac{(-1)^k}{(k+1)C^k} = C \log \dfrac{C+1}{C}$

Is it correct? What would be a proof?

Answer 1

$\sum\limits_{k=0}^\infty r^k=\frac{1}{1-r}, |r|<1$

Integrate both sides, $\sum\limits_{k=0}^\infty \frac{1}{1+k}r^{k+1}=-\log(1-r), |r|<1$

Divide r ,$\sum\limits_{k=0}^\infty \frac{1}{1+k}r^{k}=-\frac{\log(1-r)}{r}, |r|<1$

Plug $r=-\frac{1}{C}$ you get $\sum\limits_{k=0}^\infty \dfrac{(-1)^k}{(k+1)C^k} = C \log \dfrac{C+1}{C}, \forall C>1$

We can do all these things because the series is uniformly convergent in its radius of convergence.

Then we check above result for $C=1$, since the alternating harmonic series is convergent, we can plug in $r=-1$ to our series representation of $-\frac{\log(1-r)}{r}$ to get the value of the series by Abel's theorem for power series.

Hence we proved the claim.

Answer 2

Consider Taylor expansion of function $\ln(1+x)$ at $0$ and put $x=\frac{1}{C}$. The series:

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

converges when $|x|<1$, so also for $X=\frac{1}{C} < 1$.