Results 1 to 6 of 6

Math Help - Infinite Series

  1. #1
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2

    Infinite Series

    Dear friends, for \alpha \in\mathbb{R}, it appears that \displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\  right)}{\sqrt{4\alpha+1}}.

    But I'm unable to show it! Can you show it? I tried to use the identity:

     \displaystyle \frac{1}{a^2+a-b} = \frac{2}{\sqrt{4b+1}\left(2a-\sqrt{4b+1}+1\right)}-\frac{2}{\sqrt{4b+1}\left(2a+\sqrt{4b+1}+1\right)},

    and turn it into an integral, but I faced divergence issues and the whole thing evaluated to 1.
    Last edited by TheCoffeeMachine; January 26th 2011 at 02:15 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391
    Maybe you can reduce this to a telescopic series if you factorise the denominator and use partial fractions...

    \displaystyle k^2 + k - \alpha = k^2 + k + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - \alpha

    \displaystyle = \left(k + \frac{1}{2}\right)^2 - \left(\frac{4\alpha + 1}{4}\right)

    \displaystyle = \left(k + \frac{1}{2}\right)^2 - \left(\frac{\sqrt{4\alpha + 1}}{2}\right)^2

    \displaystyle = \left(k + \frac{1}{2} - \frac{\sqrt{4\alpha + 1}}{2}\right)\left(k + \frac{1}{2} + \frac{\sqrt{4\alpha + 1}}{2}\right)

    \displaystyle = \left(k + \frac{1 - \sqrt{4\alpha + 1}}{2}\right)\left(k + \frac{1 + \sqrt{4\alpha + 1}}{2}\right)...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Prove It -- The identity I had in the first post gives the partial fraction form of the summand:

    \displaystyle \frac{1}{k^2+k-\alpha} = \frac{2}{\sqrt{4\alpha+1}\left(2k-\sqrt{4\alpha+1}+1\right)}-\frac{2}{\sqrt{4\alpha+1}\left(2k+\sqrt{4\alpha+1}  +1\right)}.

    I wouldn't expect it to telescope to something simple; the answer is complicated enough, obviously.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    A way to solve this problem starts defining the function...

    \displaystyle \psi(x)= \sum_{k=1}^{\infty} \frac{x}{k\ (k+x)} + c (1)

    ... where c is a constant that for the moment we left undefined. From (1) we derive the 'fundamental identity'...

    \displaystyle \psi(1+x) - \psi(x)= \sum_{k=1}^{\infty} \{\frac{1+x}{k\ (k+1+x)}- \frac{x}{k\ (k+x)}\} =

    \displaystyle = \sum_{k=1}^{\infty} \{\frac{1}{k+x)}- \frac{1}{k+1+x)}\} = \frac{1}{1+x} \implies

    \displaystyle \implies \psi(1+x)= \psi(x) + \frac{1}{1+x} (2)

    A first important result we obtain from (2) is that we are able to valuate the sum...

    \displaystyle \sigma(t)= \sum_{k=1}^{n} \frac{1}{k+t} (3)

    Setting in (2) x=k+t-1 we obtain...

    \displaystyle \psi(k+t)-\psi(k+t-1)= \frac{1}{k+t} \implies

    \displaystyle \implies \sum_{k=1}^{n} \frac{1}{k+t}= \sum_{k=1}^{n} \{\psi(k+t) - \psi(k+t-1)\}= \psi(n+t)-\psi(t) (4)

    A second important result is that we are able to compute in terms of \psi(*) the infinite sum...

    \displaystyle S= \sum_{k=1}^{\infty} \frac{1}{(k+a)\ (k+b)} (5)

    ... and that will be explained in a successive post. A problem however is still unsolved: what is \psi(*)?... the answer starts from the definition of the 'factorial function' ...

    \displaystyle x!= \int_{0}^{\infty} t^{x}\ e^{-t}\ dt (6)

    Integrating by parts (6) we arrive to the well known identity...

    \displaystyle (1+x)!= x!\ (1+x) (7)

    ... and deriving (7)...

    \displaystyle \{(1+x)!\}^{'} = x!+ \{x!\}^{'}\ (1+x) \implies \frac{\{(1+x)!\}^{'}}{(1+x)!} = \frac{\{x!\}^{'}}{x!} + \frac{1}{1+x} (8)

    Comparing (8) and (2) we arrive to conlude that is...

    \displaystyle \psi(x)= \frac{d}{dx} \ln (x!) (9)

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Now we use the results of the previous post to valuate the series...

    \displaystyle S=\sum_{k=1}^{\infty} \frac{1}{(k+a)\ (k+b)} (1)

    Setting \sigma_{n} the n-th partial sum of (1) and taking into account the formula (4) of the previous post we have...

    \displaystyle \sigma_{n}= \sum_{k=1}^{n} \frac{1}{(k+a)\ (k+b)}= \frac{1}{b-a} \sum_{k=1}^{n} (\frac{1}{k+a}- \frac{1}{k+b})=

    \displaystyle = \frac{1}{b-a} \{\psi(n+a)-\psi(a) - \psi(n+b) + \psi(b)\} (2)

    At this point if in the previous post we set in the formula (1) c=-\gamma, where \gamma is the Euler's constant, and in the formula (2) x=k we obtain...

    \displaystyle \sum_{k=0}^{n-1} \frac{1}{k+1}=  \sum_{k=0}^{n-1} \{\psi(k+1)- \psi(k)\} = \psi(n)-\psi(0) = \psi(n) + \gamma \implies

    \displaystyle \implies \psi(n)= \sum_{k=1}^{n} \frac{1}{k} - \gamma \sim \ln n (3)

    But consequence of (3) is that...

    \displaystyle \lim_{n \rightarrow \infty} \psi(n+a)- \psi(n+b) =0 (4)

    ... so that (1) and (2) become...

    \displaystyle S=\sum_{k=1}^{\infty} \frac{1}{(k+a)\ (k+b)}= \frac{\psi(b)-\psi(a)}{b-a} (5)

    ... where...

    \displaystyle \psi(x)= \frac{d}{dx} \ln (x!) (6)

    At this point we have [I do hope...] all the elements to solve the problem proposed by TheCoffeMachine...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Nice job, Chisigma! Many thanks.
    Two of the most epic posts in the forum right there!
    I must say I didn't expect it to be that much involved.
    I'll spend some time to understand this digamma stuff!
    Last edited by TheCoffeeMachine; January 27th 2011 at 05:14 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Fourier series to calculate an infinite series
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 4th 2010, 01:49 PM
  2. Replies: 7
    Last Post: October 12th 2009, 10:10 AM
  3. Replies: 2
    Last Post: September 16th 2009, 07:56 AM
  4. infinite series and power series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 13th 2009, 04:21 AM
  5. Replies: 1
    Last Post: May 5th 2008, 09:44 PM

Search Tags


/mathhelpforum @mathhelpforum