Justify the equation $\sum_{n=0}^{\infty} \frac{(-1)^n}{m+nk} = \int_{0}^1 \frac{t^{m-1}}{1+t^k} dt$, $m > 0$.

Question

I am trying to solve this problem from my exam prep under real analytic function section.

Justify the equation $\sum_{n=0}^{\infty} \frac{(-1)^n}{m+nk} = \int_{0}^1 \frac{t^{m-1}}{1+t^k} dt$, $m > 0$.

I am not sure what $k$ is as the question doesn't really specify.

Any help will be appreciated.

Answer 1

HINT:

$$\frac1{m+nk} =\int_0^1 t^{m+nk-1}\,dt $$

Answer 2

Expanding $\frac{1}{1+t^k}$ as a geometric series in $(0,1)$ gives

$$\int^1_0\frac{t^{m-1}}{1+t^k}\,dt=\int^1_0\sum_{n\geq0}(-1)^nt^{m-1+nk}\,dt= \sum_{n\geq0}(-1)^n\int^1_0t^{m-1+nk}\,dt=\sum_{n\geq0}\frac{(-1)^n}{m+nk}$$

The change of the order of integration and summation can be justified by either Fubini's theorem of by dominated convergence.