Finding the value to which a series converge.

Question

The problem is the following:

I found, in a previous exercise, that $$\sum_0^\infty{x^k}=\frac{1}{1-x}, \space for\space|x|<1$$, and in the following exercise, it asks me to apply that result to solve the following: $$\sum_0^\infty{(-1)^k}{e^{-sk}}$$ where $s>0.$ My attempt at that was: $$\sum_0^\infty{(-1)^k}{e^{-sk}}=1-e^{-s}+e^{-2s}-e^{-3s}+...=\sum_0^\infty{e^{-2sk}}-\sum_0^\infty{e^{-s(2k+1)}}\\=\sum_0^\infty{e^{-2sk}}-e^{-s}\sum_0^\infty{e^{-s(2k)}}=\frac{1}{1-e^{-2s}}-\frac{e^{-s}}{1-e^{-2s}}$$, but the answer the book gave was $$\frac{1}{1+e^{-s}}$$, however I couldn't rearrage my answer to match that. Is something wrong before I get to the answer or is this right and the only problem is my difficulty in rearraging the terms?

Answer 1

Both answers are correct. We have $$\frac{1}{1-e^{-2s}}-\frac{e^{-s}}{1-e^{-2s}} = \frac{1- e^{-s}}{1-e^{-2s}} =\frac{1- e^{-s}}{(1-e^{-s})(1 + e^{-s})} = \frac{1}{1 + e^{-s}}. $$

Answer 2

While the other answer and comment help you see that your answer is still correct, maybe it is also worth pointing out that the exercise is expecting you to recognize the corresponding $x$ and leverage the knowledge you have from the previous exercise more directly instead of going through any more pain than needed. That is, the exercise expects you to see

$$\sum_{k=0}^\infty{(-1)^k}{e^{-sk}} = \sum_{k=0}^\infty\left[(-1)e^{-s}\right]^{k}$$ which means you can set $x=(-1)e^{-s}$ in the result from the previous exercise and the answer is immediate. Note that for $s >0$, we have $e^{-s} < 1$ so that the previous exercise's condition $|x|<1$ is satisfied.