View topic - series summation • Art of Problem Solving

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Author	Message
misan Posts: 1107 Location: Mountain View, California	Posted: Jun 21, 2007, 11:57 pm • # 1 find the sum $\sum_{k=0}^\infty\;\;\frac{(-1)^{k}}{5k+1}$

Carcul Posts: 4018 Location: Pampilhosa	Posted: Jun 22, 2007, 1:19 am • # 2 $\sum_{k=0}^{\infty}\frac{(-1)^{k}}{5k+1}= \int^{1}_{0}\frac{1}{1+x^{5}}\,dx$ .

Kent Merryfield

Posts: 12103
Location: Long Beach, CA
United States

Posted: Jun 22, 2007, 7:38 am • # 3

My thoughts on seeing $\int_{0}^{1}\frac{dx}{1+x^{5}}$ :

This is a rational function. Can we find, in closed form, the factors (linear and quadratic) of the denominator? Yes we can; I do know how to do that (lots of golden means and other uses of $\sqrt{5}$ ). If you can factor the denominator, then, through partial fractions, you can compute the integral in closed form.

Therefore it is possible to compute this integral in closed form. Would it be possible to do that by hand? Yes, it would. Am I personally willing to demonstrate that? Not today; maybe not ever.

It's time for your favorite computer algebra system (Maple, Mathematica, Derive, TI-92 or TI-89, ... )

Last edited by Kent Merryfield on Jun 22, 2007, 9:01 am, edited 1 time in total.

sylow_theory

Posts: 902

Posted: Jun 22, 2007, 8:55 am • # 4

Kent Merryfield wrote:

It's time for your favorite computer algebra system (Maple, Mathematica, Derive, TI-92 or TI-89, ... )

I once wondered about the integral,
$\int \frac{1}{1+x^{n}}dx$
Whether it has closed form.
After using the "Integrator" for many different values I am convinced yes.
However, the size of the anti-derivative gross exponentially .

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TZF

Posts: 3276
Location: Ithaca, New York
United States

Blog: View Blog

Posted: Jun 22, 2007, 10:22 am • # 5

This is what my favorite computer algebra system (Mathematica 6) says:

$\int^{1}_{0}\frac{1}{1+x^{n}}dx =\frac{\frac{\Gamma ' \left( \frac{n+1}{2n}\right)}{\Gamma \left( \frac{n+1}{2n}\right)}-\fra...$

If I try an indefinite integral, it's giving it in terms of some HyperGeometric2F1 function...
$\int = x \left( \text{HyperGeometric2F1}\left(\frac{1}{n},1,1+\frac{1}{n},x^{-n}\right) \right)+C$
Looking in the help menu, HyperGeometric2F1 is series-defined, so that really doesn't help.

As for the original problem, the first formula above works; it's:

$S=-\frac{1+\sqrt{5}}{80}\left(-8 \sqrt{\frac{2}{\sqrt{5}+5}}\pi+2(\sqrt{5}-3) \text{ArcCsch}(2)-5 \sqrt{5}\ln 2+\ln 8+\ln \le...$

$S \approx 0.8883135726516996 \ldots$

after using FullSimplify.

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misan Posts: 1107 Location: Mountain View, California	Posted: Jun 23, 2007, 1:58 am • # 6 The Zuton Force wrote: This is what my favorite computer algebra system (Mathematica 6) says: $S=-\frac{1+\sqrt{5}}{80}\left(-8 \sqrt{\frac{2}{\sqrt{5}+5}}\pi+2(\sqrt{5}-3) \text{ArcCsch}(2)-5 \sqrt{5}\ln 2+\ln 8+\ln \le...$ $S \approx 0.8883135726516996 \ldots$ after using FullSimplify. holy smokes !!

Doe John

Posts: 151

Posted: Apr 15, 2008, 3:33 pm • # 7

$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}=\frac{\pi}{4} \csc\left(\frac{\pi}{2}\right)$

$\sum_{k=0}^\infty \frac{(-1)^k}{3k+1}=\frac{\pi}{6} \csc\left(\frac{\pi}{3}\right)+\frac{\ln 2}{3}$

$\sum_{k=0}^\infty \frac{(-1)^k}{4k+1}=\frac{\pi}{8} \csc\left(\frac{\pi}{4}\right)+\frac{\ln(1+\sqrt{2})}{2\sqrt{2}}$

$\sum_{k=0}^\infty \frac{(-1)^k}{5k+1}=\frac{\pi}{10}\csc\left(\frac{\pi}{5}\right)+\frac{1}{\sqrt{5}}\ln\left(\frac{1+\sqrt{5...$

$\sum_{k=0}^\infty \frac{(-1)^k}{6k+1}=\frac{\pi}{12} \csc\left(\frac{\pi}{6}\right)+\frac{\ln(2+\sqrt{3})}{2\sqrt{3}}$

jmerry Posts: 7825 Location: Seattle	Posted: Apr 15, 2008, 4:01 pm • # 8 Now, the sum from $-\infty$ to $\infty$ is actually nice- and I posted it here a few years ago.

Blouge

Posts: 9

Posted: Apr 17, 2008, 8:50 pm • # 9

Using the fact that Digamma(s+1) = Digamma(s) + 1/s, we can express these series in terms of the digamma function:

10/6 + 10/16 + 10/26 + ... = Digamma(infinity) - Digamma(6/10)
10/11 +10/21 + 10/31 + ... = Digamma(infinity) - Digamma(11/10)

Hence, the sum we want is 1 - (1/6 - 1/11 + 1/16 -_... = 1 - (Digamma(11/10) - Digamma(6/10)) / 10.

Gauss's formula for the Digamma function at rational numbers can be used to calculate explicit values for these digamma expressions.
As posted earlier, the result involves sin and cos of (pi/10) and can be expressed in terms of logarithms, radicals, the gamma constant, and pi.
(Note: cos(pi/5) = (1+sqrt(5))/4).

Edit: I forgot the initial 1- in the sum (for index k = 0).

Doe John

Posts: 151

Posted: Apr 21, 2008, 1:03 am • # 10

$S = \sum_{k = 0}^\infty \frac {( - 1)^k}{5k + 1} = \frac {1}{10} \left[\psi\left(\frac {6}{10}\right) - \psi\left(\frac {1}{1...$

After applying the digamma theorem

$\psi\left(\frac {p}{q}\right) = - \gamma - \ln(2q) - \frac {\pi}{2}\cot\left(\frac {p\pi}{q}\right) + \sum_{k = 1}^{q - 1} \c...$

$S = \frac {1}{10} \left[\frac {\pi}{2}\left(\cot \frac {\pi}{10} - \cot \frac {6\pi}{10}\right) + \sum_{k = 1}^9 \left[\cos\l...$

$= \frac {1}{10} \left[\pi \csc \frac {\pi}{5} - \frac {\sqrt {5} + 1}{2}\ln \frac {\sqrt {5} - 1}{4} + 0 + \frac {\sqrt {5} -...$

Hence $S - \frac {\pi}{10} \csc\frac {\pi}{5} = \frac15\left[\frac {\sqrt {5} - 1}{2}\ln \frac {\sqrt {5} + 1}{4} - \frac {\sqrt {5}...$

$= \frac15 \left[\frac {\sqrt {5}}{2}\left[\ln \frac {\sqrt {5} + 1}{4} - \ln \frac {\sqrt {5} - 1}{4}\right] - \frac12\left[\...$

$= \frac15 \left[\frac {\sqrt {5}}{2}\ln\left(\frac {\sqrt {5} + 1}{2}\right)^2 - \frac12 \ln \frac14\right] = \frac {1}{\sqrt...$

Extremal Posts: 1391 Location: Saint-Petersburg	Posted: Apr 21, 2008, 6:26 am • # 11 Doe John : Do you know any article where can I find digamma theorem ? I think it's can be proved by residue , _________________ I believe ...

Doe John Posts: 151	Posted: Apr 22, 2008, 9:50 am • # 12 You find a proof of Gauss' digamma theorem on PlanetMath.

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