# Expand $f(x) = \sum_{k=0}^{+\infty} \frac{a^k}{(1+(a^kx)^2)k!}$, $g(x)=\sum_{k=0}^{+\infty} \frac{(-1)^ka^k}{(1+(a^kx)^2)k!}$ into power series.

As stated in the title, expand $f(x) = \sum_{k=0}^{+\infty} \frac{a^k}{(1+(a^kx)^2)k!}$, $g(x)=\sum_{k=0}^{+\infty} \frac{(-1)^ka^k}{(1+(a^kx)^2)k!}$ into power series, where $0<a<1$ and $x \in (-1, 1)$.

So far my only ideas are that the functions which we're summing are derivatives of $F_{k}(x)=\frac{\arctan(a^kx)}{k!}$, $G_{k}(x)=\frac{\arctan(a^kx)}{k!}$, which I don't really know how to use.

Also, another idea, which looks very complicated: $f(0)=e^a$, $g(0)=e^{-a}$, so maybe it's possible to expand $f, g$ into Taylor series.

Expand geometricaly \begin{eqnarray*} f(a,x)= \sum_{k=0}^{\infty} \frac{a^k}{1+(a^kx)^2} =\sum_{k=0}^{\infty} \sum_{j=0}^{\infty} (-1)^j a^k (a^kx)^{2j} \end{eqnarray*} now interchange the order of the sums & perform the inner geometric sum \begin{eqnarray*} f(a,x)= \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} (-1)^j a^{k(2j+1)} x^{2j}=\sum_{j=0}^{\infty} \frac{(-1)^j x^{2j}}{1-a^{2j+1}}. \end{eqnarray*} The other sum can be done similarly & will have a $+$ in the denominator.
• How can $j$ be in the result when it is an index of summation? – marty cohen May 24 '17 at 23:45
• @martycohen I have edited now ... you are right I had managed to loose the $j$ sum as well as the $k$ sum ! ... thanks for pointing this out ... $\ddot \smile$ – Donald Splutterwit May 25 '17 at 10:38