Originally posted by DeepThought
So:
1/(N^2(N+1)) = 1/N^2 - 1/N + 1/(N + 1)
Then you either have known sums you can look up or can use a generating function to get the sum.
Thanks for that 🙂
I confess complete ignorance on the method of partial fractions.
But I find that equation extremely useful and gives me a breakthrough, of sorts.
This is because we now have:
1/(X^2(X+1)) = 1/(X^2) + 1/(X + 1) – 1/X (using the method of partial fractions )
so:
∑ [X = x, X = +∞] 1/(X^2(X+1)) = ∑ [X = x, X = +∞] ( 1/(X^2) + 1/(X + 1) – 1/X)
Note the sum of ∑ [X = x, X = +∞] 1/X
and ∑ [X = x, X = +∞] 1/(X + 1) both unhelpfully sums up to infinity.
However the sum of ∑ [X = x, X = +∞] (1/(X + 1) – 1/X ) rather helpfully sums up to simply -1/x i.e.
∑ [X = x, X = +∞] (1/(X + 1) – 1/X ) = -1/x.
So now we have:
∑ [X = x, X = +∞] 1/(X^2(X+1))
= ∑ [X = x, X = +∞] ( 1/X^2 + 1/(X + 1) – 1/X)
= ( ∑ [X = x, X = +∞] 1/X^2 ) + ∑ [X = x, X = +∞] (1/(X + 1) – 1/X)
= ( ∑ [X = x, X = +∞] 1/X^2 ) – 1/x
so now we have a desirable simplification of the sum of the infinite series we need to evaluate.
Now explaining stuff and speaking here for the benefit of
other readers that may be here who are not so knowledgeable of maths as us:
Note that, if you think about what the summation means (obviously, you would know very clearly what summation means, ∑ [X = x, X = +∞] 1/X^2 must equal what the sum would equal
∑ [X = 1, X = +∞] 1/X^2 subtract all the terms of the series of the summation from the first one in that series to the xth – 1 one in that series.
For example; if x =3, then we have:
∑ [X = 3, X = +∞] 1/X^2 = 1/(3^2) + 1/(4^2) + 1/(5^2) + 1/(6^2) ...
∑ [X = 1, X = +∞] 1/X^2 = 1/(1^2) + 1/(2^2) + 1/(3^2) + 1/(4^2) + 1/(5^2) + 1/(6^2) …
and notice most of the terms in the two above series of the summations are identical and the only difference between the two series of the summations is that the first series doesn't have the first two terms the second series does with those two terms being 1/(1^2) and 1/(2^2).
From this, we can reason that:
∑ [X = 3, X = +∞] 1/X^2 = ( ∑ [X = 1, X = +∞] 1/(X^2) ) – 1/(1^2) – 1/(2^2)
and the more general equation for any x is:
∑ [X = x, X = +∞] 1/X^2 = ( ∑ [X = 1, X = +∞] 1/(X^2) ) – ∑ [X = 1, X = x – 1] 1/(X^2)
But ∑ [X = 1, X = +∞] 1/X^2 of the above is equal to just a constant which is:
∑ [X = 1, X = +∞] 1/X^2 = (π^2)/6
So we now have:
∑ [X = x, X = +∞] 1/X^2 = ( ∑ [X = 1, X = +∞] 1/(X^2) ) – ∑ [X = 1, X = x – 1] 1/(X^2)
= ( (π^2)/6 ) – ∑ [X = 1, X = x – 1] 1/X^2
and thus:
∑ [X = x, X = +∞] 1/(X^2(X+1)) = ( (π^2)/6 ) – ( ∑ [X = 1, X = x – 1] 1/(X^2) ) – 1/x
so now we have a desirable change of the sum we need to evaluate from being of an infinite series to a more manageable finite series of ∑ [X = 1, X = x – 1] 1/(X^2).
And that ∑ [X = 1, X = x – 1] 1/(X^2) sum is a harmonic series (I confess I know next to nothing about the mathematics of harmonic series. So not sure where I can go from this point. But I still like the look of this! )