# Find the limits of "Almost Divergent" Series

Find the following limits: $\lim_{\varepsilon\rightarrow 0}\sum_{n=0}^{+\infty}\frac{(-1)^n}{1+n\epsilon}$

$\lim_{\varepsilon\rightarrow 0}\sum_{n=0}^{+\infty}\frac{(-1)^n}{1+n^{2}\epsilon}$

• Interesting question..! Are you sure the first one converges? Commented Jan 21, 2014 at 16:43

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.

• Note that this approach seems to work for $\sum_{n=0}^{\infty}(-1)^{n}/(1+n^k\varepsilon)$ for any $k$, and it always gives $1/2$ as $\varepsilon\rightarrow 0$. Commented Jan 21, 2014 at 16:58
• Well done,thank you.But see here for more challenging ones. Commented Jan 23, 2014 at 11:16

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.

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 :)

• Well done,thank you.But see here for more challenging ones.:) Commented Jan 23, 2014 at 11:18

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.