Vtyjkyuk

I just proved $sumfracx^n(2n)!$ is convergent for all $x in mathbb R$. Define the function $f: mathbb R rightarrow mathbb R$ by
$$f(x) = sum_n=0^infty fracx^n(2n)!$$
Now the question says; prove that the series $sumfrac1f(k)$ is convergent.

Now I'm having difficulty understanding this question, do we sum up like $sum_k=0^inftyfrac1f(k)$ hence having to kinda deal with $f: mathbb N rightarrow mathbb R$ or is it something different? (I'm suspecting that to be honest, I mean what else could it really be).

Then after clarification, hints are preferred as this is test preparation :)

For $xgeq 1$, $$frace^x+e^-x2=sum_n=0^inftyfracx^2n(2n)!.$$
So $$frace^sqrtx+e^-sqrtx2=sum_n=0^inftyfracx^n(2n)!.$$
So $$f(k)=frace^sqrtk+e^-sqrtk2,kgeq 1,$$
and the series $$sum frac1f(k)$$ is convergent.

You can solve this without explicitly knowing what $f(k)$ is. Clearly you have for $k>0$ that $f(k)ge frack^24!$ so $frac1f(k)le frac24k^2$. Now you can use the comparison test and the fact that $sum_k=1^infty frac1k^2$ converges.

Yes, you sum up as $;sum_k=0^infty f(k)$. Explicitly, this is a double sum:
$$sum_k=0^infty Biggl(frac 1sum_n=0^infty frack^n(2n)!Biggr).$$

Hint:
$sum_k=0^+infty x^n/(2n)! = cosh(sqrtx)$ and $frac1coshsqrtx = operatornamesech sqrtx$ then use comparison test.