Find the limits of "Almost Divergent" Series Find the following limits:
$ \lim_{\varepsilon\rightarrow 0}\sum_{n=0}^{+\infty}\frac{(-1)^n}{1+n\epsilon} $
$ \lim_{\varepsilon\rightarrow 0}\sum_{n=0}^{+\infty}\frac{(-1)^n}{1+n^{2}\epsilon} $
 A: Note that
$$
\sum_{n=0}^{\infty}(-1)^{n}a_n=\sum_{n=0}^{\infty}(a_{2n}-a_{2n+1}).
$$
If $a_n=1/(1+n\varepsilon)$, then
$$
a_{2n}-a_{2n+1}=\frac{1}{1+2n\varepsilon} - \frac{1}{1+(2n+1)\varepsilon}=\frac{\varepsilon}{(1+2n\varepsilon)(1+(2n+1)\varepsilon)}.
$$
Writing $x=n\varepsilon$, we can replace the sum by an integral:
$$
\sum_{n=0}^{\infty}\frac{(-1)^{n}}{1+n\varepsilon}= \sum_{x=0,\varepsilon,2\varepsilon,\ldots}\frac{\varepsilon}{(1+2x)(1+2x+\varepsilon)}\sim\int_{0}^{\infty}\frac{dx}{(1+2x)^2}=\frac{1}{2}.
$$
Similarly, if $a_n=1/(1+n^2\varepsilon)$, then
$$
a_{2n}-a_{2n+1}=\frac{1}{1+(2n)^2\varepsilon} - \frac{1}{1+(2n+1)^2\varepsilon}\sim\frac{4n\varepsilon}{(1+4n^2\varepsilon)^2}.
$$
Here we take $x=n\sqrt{\varepsilon}$, giving us
$$
\sum_{n=0}^{\infty}\frac{(-1)^{n}}{1+n^2\varepsilon}=\sum_{x=0,\sqrt{\varepsilon},2\sqrt{\varepsilon},\ldots}\frac{4x \sqrt{\varepsilon}}{(1+4x^2)^2}\sim\int_{0}^{\infty}\frac{4xdx}{(1+4x^2)^2}=\frac{1}{2}
$$
as well.
A: I do not know how this could help. Your first summation corresponds almost to the definition of the Hurwitz-Lerch transcendent function since the result is 
HurwitzLerchPhi[-1, 1, 1 / epsilon] / epsilon  
Using the limits, when $\epsilon\ ->0$, the limit is simply 1/2.
A: For the first one we could write
$$
\frac{1}{1+e^{-\epsilon x}} = \sum_{n=0}^{\infty} (-1)^n e^{-\epsilon n x}
$$
and integrate term-by-term to find that
$$
\sum_{n=0}^{\infty} \frac{(-1)^n}{1+\epsilon n} = \int_0^{\infty} \frac{e^{-x}}{1+e^{-\epsilon x}}\,dx.
$$
By the dominated convergence theorem we then have
$$
\lim_{\epsilon \to 0} \int_0^{\infty} \frac{e^{-x}}{1+e^{-\epsilon x}}\,dx = \frac{1}{2} \int_0^\infty e^{-x}\,dx = \frac{1}{2}.
$$
The second one resists this approach :)
A: Using Claude Leibovici's idea of using Mathematica, I get the following
$$
\sum_{n=0}^{+\infty}\frac{(-1)^n}{1+n\epsilon} = \frac{\Phi \left(-1,1,\frac{1}{\epsilon }\right)}{\epsilon },
$$
where $\Phi$ is the Hurwitz-Lurch trancendent function. The limit of the above expression is indeed $1/2$ as claimed. For your other function I get
$$
\sum_{n=0}^{+\infty}\frac{(-1)^n}{1+n^{2}\epsilon} = \frac{2 \sqrt{\epsilon }+\pi  \coth \left(\frac{\pi }{2 \sqrt{\epsilon }}\right)}{4 \sqrt{\epsilon }}-\frac{\pi  \tanh \left(\frac{\pi }{2 \sqrt{\epsilon }}\right)}{4 \sqrt{\epsilon }}
$$
This also has limit $1/2$.
The identities used in the derivations are all from Mathematica. I believe the answer of $1/2$ would also be reached by the Cesaro and Abel summations on the divergent series.
