# Thread: Infinite Series

1. ## Infinite Series

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

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

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

and turn it into an integral, but I faced divergence issues and the whole thing evaluated to $\displaystyle 1$.

2. Maybe you can reduce this to a telescopic series if you factorise the denominator and use partial fractions...

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

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

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

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

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

3. Prove It -- The identity I had in the first post gives the partial fraction form of the summand:

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

I wouldn't expect it to telescope to something simple; the answer is complicated enough, obviously.

4. A way to solve this problem starts defining the function...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

... and deriving (7)...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Now we use the results of the previous post to valuate the series...

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

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

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

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

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

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

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

But consequence of (3) is that...

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

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

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

... where...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. 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!