-1/12
Let’s define three different series
To find
Let’s place a shifted copy of itself and add the corresponding terms 
Now lets find 
The second proof also makes use of the idea we learned that

The way he goes to proof is using the idea of power series and differentiating it

Differentiating both sides we get

Now plug in
on both sides we get
and we have the same result.
The next thing he does is invoke the Reimann zeta function
Notice that
Now multiply both sides of
by
we get

Now subtracting this equation from the original
we get 
Plugging back
we get




The way he goes to proof is using the idea of power series and differentiating it
Differentiating both sides we get
Now plug in
The next thing he does is invoke the Reimann zeta function
Notice that
Now multiply both sides of
Now subtracting this equation from the original
Plugging back
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