Maybe you can reduce this to a telescopic series if you factorise the denominator and use partial fractions...
...
Dear friends, for , it appears that .
But I'm unable to show it! Can you show it? I tried to use the identity:
,
and turn it into an integral, but I faced divergence issues and the whole thing evaluated to .
A way to solve this problem starts defining the function...
(1)
... where c is a constant that for the moment we left undefined. From (1) we derive the 'fundamental identity'...
(2)
A first important result we obtain from (2) is that we are able to valuate the sum...
(3)
Setting in (2) we obtain...
(4)
A second important result is that we are able to compute in terms of the infinite sum...
(5)
... and that will be explained in a successive post. A problem however is still unsolved: what is ?... the answer starts from the definition of the 'factorial function' ...
(6)
Integrating by parts (6) we arrive to the well known identity...
(7)
... and deriving (7)...
(8)
Comparing (8) and (2) we arrive to conlude that is...
(9)
Kind regards
Now we use the results of the previous post to valuate the series...
(1)
Setting the n-th partial sum of (1) and taking into account the formula (4) of the previous post we have...
(2)
At this point if in the previous post we set in the formula (1) , where is the Euler's constant, and in the formula (2) we obtain...
(3)
But consequence of (3) is that...
(4)
... so that (1) and (2) become...
(5)
... where...
(6)
At this point we have [I do hope...] all the elements to solve the problem proposed by TheCoffeMachine...
Kind regards
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!