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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} $

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  • $\begingroup$ Interesting question..! Are you sure the first one converges? $\endgroup$
    – user76568
    Commented Jan 21, 2014 at 16:43

4 Answers 4

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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.

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    $\begingroup$ 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 $\varepsilon\rightarrow 0$. $\endgroup$
    – mjqxxxx
    Commented Jan 21, 2014 at 16:58
  • $\begingroup$ Well done,thank you.But see here for more challenging ones. $\endgroup$
    – Lagrenge
    Commented Jan 23, 2014 at 11:16
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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.

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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 :)

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  • $\begingroup$ Well done,thank you.But see here for more challenging ones.:) $\endgroup$
    – Lagrenge
    Commented Jan 23, 2014 at 11:18
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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.

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