Thread: Is there a closed form solution to this infinite series?

1. Is there a closed form solution to this infinite series?

Is there a closed form solution to this infinite series?

x is a number between 0 and 1.
n can take any value between 1 and infinity.

1/n + x/(n+2) + x^2/(n+4) + x^3/(n+6) + ...................

Thanks!

2. Originally Posted by Titian
Is there a closed form solution to this infinite series?

x is a number between 0 and 1.
n can take any value between 1 and infinity.

1/n + x/(n+2) + x^2/(n+4) + x^3/(n+6) + ...................

Thanks!
Wolfram Alpha

Lerch Transcendent -- from Wolfram MathWorld

CB

3. Originally Posted by Titian
Is there a closed form solution to this infinite series?

x is a number between 0 and 1.
n can take any value between 1 and infinity.

1/n + x/(n+2) + x^2/(n+4) + x^3/(n+6) + ...................

Thanks!
Let's suppose that n is a 'natural number' so that we have a family of functions defined as...

$\displaystyle \varphi_{n} (x) = \sum_{k=0}^{\infty} \frac{x^{k}}{n+2 k}$ (1)

First we set $\xi= \sqrt{x}$ and then we start with n=1…

$\displaystyle \varphi_{1} (\xi)= 1 + \frac{\xi^{2}}{3} + \frac{\xi^{4}}{5} + \frac{\xi^{6}}{7} + ... = \frac{1}{\xi}\ (\xi + \frac{\xi^{3}}{3} + \frac{\xi^{5}}{5} + \frac{\xi^{7}}{7} + ...) =$

$\displaystyle = \frac{1}{\xi}\ \frac{\ln (1+\xi)- \ln (1-\xi)}{2} = \frac{1}{2 \xi}\ \ln \frac{1+\xi}{1-\xi}$ (1)

Now for n=2...

$\displaystyle \varphi_{2} (\xi)= \frac{1}{2} + \frac{\xi^{2}}{4} + \frac{\xi^{4}}{6} + \frac{\xi^{6}}{8} +...= \frac{1}{\xi^{2}}\ (\frac{\xi^{2}}{2} + \frac{\xi^{4}}{4} + \frac{\xi^{6}}{6} + \frac{\xi^{8}}{8}+ ...)=$

$\displaystyle = \frac{1}{\xi^{2}}\ \frac{- \ln (1+\xi) - \ln (1-\xi)}{2} = \frac{1}{2 \xi^{2}} \ \ln \frac{1}{(1+\xi)\ (1-\xi)}$ (2)

Now for n=3...

$\displaystyle \varphi_{3} (\xi)= \frac{1}{3} + \frac{\xi^{2}}{5} + \frac{\xi^{4}}{7} + \frac{\xi^{6}}{9} +...= \frac{1}{\xi^{3}}\ (\frac{\xi^{3}}{3} + \frac{\xi^{5}}{5} + \frac{\xi^{7}}{7} + \frac{\xi^{9}}{9}+ ...)=$

$\displaystyle = \frac{1}{\xi^{3}}\ \{\frac{\ln (1+\xi) - \ln (1-\xi)}{2} -\xi\} = \frac{1}{2 \xi^{3}}\ \ln \frac{1+\xi}{1-\xi} - \frac{1}{\xi^{2}}$ (3)

And now for n=4...

$\displaystyle \varphi_{4} (\xi)= \frac{1}{4} + \frac{\xi^{2}}{6} + \frac{\xi^{4}}{8} + \frac{\xi^{6}}{10} +...= \frac{1}{\xi^{4}}\ (\frac{\xi^{4}}{4} + \frac{\xi^{6}}{6} + \frac{\xi^{8}}{8} + \frac{\xi^{10}}{10}+ ...)=$

$\displaystyle = \frac{1}{\xi^{4}}\ \{\frac{- \ln (1+\xi) - \ln (1-\xi)}{2} -\frac{\xi^{2}}{2} \} = \frac{1}{2 \xi^{4}}\ \{\ln \frac{1}{(1+\xi)\ (1-\xi)}- \frac{1}{\xi^{2}}\}$ (4)

Observing (1), (2), (3) and (4) it seems that the general explicit expression for $\varphi_{n} (x)$ is...

$\displaystyle \varphi_{n}(x)=\left\{\begin{array}{ll} x^{-\frac{n}{2}}\ \{\frac{1}{2}\ \ln \frac{1}{(1+x^{\frac{1}{2}})\ (1-x^{\frac{1}{2}})} - \sum_{k=1}^{\frac{n}{2}-1} \frac{x^{k}}{2 k} \} ,\,\, n\ even \\{}\\x^{-\frac{n}{2}}\ \{\frac{1}{2}\ \ln \frac{1+ x^{\frac{1}{2}}}{1-x^{\frac{1}{2}}} - \sum_{k=1}^{\frac{n-1}{2}} \frac{x^{k -\frac{1}{2}}}{2 k-1} \} ,\,\, n\ odd\end{array}\right.$ (5)

Kind regards

$\chi$ $\sigma$

P.S. : also the expressions like $\displaystyle \sum_{k} \frac{x^{k}}{2 k}$ can be written as functions of x...

4. Originally Posted by chisigma
Let's suppose that n is a 'natural number' so that we have a family of functions defined as...

$\displaystyle \varphi_{n} (x) = \sum_{k=0}^{\infty} \frac{x^{k}}{n+2 k}$ (1)

First we set $\xi= \sqrt{x}$ and then we start with n=1…

$\displaystyle \varphi_{1} (\xi)= 1 + \frac{\xi^{2}}{3} + \frac{\xi^{4}}{5} + \frac{\xi^{6}}{7} + ... = \frac{1}{\xi}\ (\xi + \frac{\xi^{3}}{3} + \frac{\xi^{5}}{5} + \frac{\xi^{7}}{7} + ...) =$

$\displaystyle = \frac{1}{\xi}\ \frac{\ln (1+\xi)- \ln (1-\xi)}{2} = \frac{1}{2 \xi}\ \ln \frac{1+\xi}{1-\xi}$ (1)

Now for n=2...

$\displaystyle \varphi_{2} (\xi)= \frac{1}{2} + \frac{\xi^{2}}{4} + \frac{\xi^{4}}{6} + \frac{\xi^{6}}{8} +...= \frac{1}{\xi^{2}}\ (\frac{\xi^{2}}{2} + \frac{\xi^{4}}{4} + \frac{\xi^{6}}{6} + \frac{\xi^{8}}{8}+ ...)=$

$\displaystyle = \frac{1}{\xi^{2}}\ \frac{- \ln (1+\xi) - \ln (1-\xi)}{2} = \frac{1}{2 \xi^{2}} \ \ln \frac{1}{(1+\xi)\ (1-\xi)}$ (2)

Now for n=3...

$\displaystyle \varphi_{3} (\xi)= \frac{1}{3} + \frac{\xi^{2}}{5} + \frac{\xi^{4}}{7} + \frac{\xi^{6}}{9} +...= \frac{1}{\xi^{3}}\ (\frac{\xi^{3}}{3} + \frac{\xi^{5}}{5} + \frac{\xi^{7}}{7} + \frac{\xi^{9}}{9}+ ...)=$

$\displaystyle = \frac{1}{\xi^{3}}\ \{\frac{\ln (1+\xi) - \ln (1-\xi)}{2} -\xi\} = \frac{1}{2 \xi^{3}}\ \ln \frac{1+\xi}{1-\xi} - \frac{1}{\xi^{2}}$ (3)

And now for n=4...

$\displaystyle \varphi_{4} (\xi)= \frac{1}{4} + \frac{\xi^{2}}{6} + \frac{\xi^{4}}{8} + \frac{\xi^{6}}{10} +...= \frac{1}{\xi^{4}}\ (\frac{\xi^{4}}{4} + \frac{\xi^{6}}{6} + \frac{\xi^{8}}{8} + \frac{\xi^{10}}{10}+ ...)=$

$\displaystyle = \frac{1}{\xi^{4}}\ \{\frac{- \ln (1+\xi) - \ln (1-\xi)}{2} -\frac{\xi^{2}}{2} \} = \frac{1}{2 \xi^{4}}\ \{\ln \frac{1}{(1+\xi)\ (1-\xi)}- \frac{1}{\xi^{2}}\}$ (4)

Observing (1), (2), (3) and (4) it seems that the general explicit expression for $\varphi_{n} (x)$ is...

$\displaystyle \varphi_{n}(x)=\left\{\begin{array}{ll} x^{-\frac{n}{2}}\ \{\frac{1}{2}\ \ln \frac{1}{(1+x^{\frac{1}{2}})\ (1-x^{\frac{1}{2}})} - \sum_{k=1}^{\frac{n}{2}-1} \frac{x^{k}}{2 k} \} ,\,\, n\ even \\{}\\x^{-\frac{n}{2}}\ \{\frac{1}{2}\ \ln \frac{1+ x^{\frac{1}{2}}}{1-x^{\frac{1}{2}}} - \sum_{k=1}^{\frac{n-1}{2}} \frac{x^{k -\frac{1}{2}}}{2 k-1} \} ,\,\, n\ odd\end{array}\right.$ (5)

Kind regards

$\chi$ $\sigma$

P.S. : also the expressions like $\displaystyle \sum_{k} \frac{x^{k}}{2 k}$ can be written as functions of x...
OK I suppose the use of n indicates that the OP wants this to be an natural.

CB