I would like to show that $S(x)=\sum_{n=1}^{+\infty} (-1)^n ln (1+ \frac{1}{n(1+x^2)})$ has a power series representation in the neighborhood of zero, but I'm a bit stuck, as : 1) I can't figure a closed form of S(x) (and I reckon there isn't any). 2) My impulse would be to using the Taylor series. The k+1-th derivative of $ln (1+ \frac{1}{n(1+x^2)})$ is lower in module than $(2/n)k!/|x+i\sqrt{1+1/n}|^{k+1}$, so $|f^{(k+1)}(c) x^{k+1}/(k+1)!|\leq (2/n(k+1))|x/(x+i\sqrt{1+1/n})|^{k+1}$, but I'm not sure what to do next. If someone has any idea it will be greatly appreciated (p.n. : I am not a student and this is not an assignment !) Thanks for any help.

  • $\begingroup$ Maybe this can help by changing the form $$\mathrm{e}^{S(x)} = \frac{\prod_{k=1}^\infty 1 + \frac{1}{2k\left(1+x^2\right)} }{\prod_{k=1}^\infty 1 + \frac{1}{(2k+1)\left(1+x^2\right)}}$$ ? $\endgroup$ – Chinny84 Sep 22 '15 at 7:10
  • $\begingroup$ Nice, but they're both divergent products aren't they? $\endgroup$ – Ernest Wilde Sep 24 '15 at 6:48

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