## Evaluate the Infinite Sum of n^2/(1+n^3)

Someone recently asked for the sum of the alternating series
inf n+1 n^2
SUM (-1) -------
n=1 1 + n^3
Knopp's book on infinite series gives this closed form expression for
the series (see below), but it's interesting to notice that for n>1
we have
n^2 1 1 1 1 1
------- = --- - --- + --- - ---- + ---- - ...
1 + n^3 n n^4 n^7 n^10 n^13
so the original sum can be evaluated in terms of "eta functions",
which are the same as zeta functions except with alternating signs.
For example, eta(4) is defined as the alternating sum of the inverses
of the fourth powers
1 1 1 1
eta(4) = --- - --- + --- - --- + ...
1^4 2^4 3^4 4^4
Of course, for even-valued arguments the eta function can be expressed
in closed form as
(2^(2k-1) - 1) pi^2k
eta(2k) = -------------------- B_2k
(2k)!
where B_2k is a Bernoulli number. Thus we have eta(4) = (7/720)pi^4
and so on. Unfortunately, there is no (known) closed-form expression
for eta with odd-valued arguments, with the exception of the well-
known series eta(1) = ln(2).
In these terms the original summation can be evaluated as
1
SUM = ln(2) - --- + (1-eta(4)) - (1-eta(7)) + (1-eta(10)) - ...
2
This is convenient because the values of eta(k) are tabulated in
many reference books. Also, notice that successive terms becomes
very small because, for example, eta(40) = .99999999999909050538.
Based on the first 13 terms of this series, ending with (1-eta(40)),
the sum is 0.23956074734...
Anyway, as explained in Knopp's book, we can expand n^2/(1+n^3) in
the usual way via partial fractions to give
_ _
n^2 1 | 1 2n - 1 |
------- = - | ----- + ----------- |
1 + n^3 3 |_ 1 + n n^2 - n + 1 _|
The contribution of 1/(1+n) is obviously just (1/3)[1-ln(2)], so we
only need to evaluate the contribution of the second component of
the summand.
Knopp shows how to derive the alternating sum
_
inf n+1 2n - 1 / /3 \
SUM (-1) ----------- = PI sech( ---- PI )
n=1 n^2 - n + 1 \ 2 /
so we have the overall result
_ _ _
inf n+1 n^2 1 | / /3 \ |
SUM (-1) ------- = - | 1 - ln(2) + PI sech( ---- PI ) |
n=1 1 + n^3 3 |_ \ 2 / _|
= .239560747340741949878...
Combining this result with the previous expression in terms of the
sum of 1-eta functions, we have the interesting result
_
inf n+1 1 1 PI / /3 \
SUM (-1) [1 - eta(3n+1)] = - + - ln(2) - -- sech( --- PI )
n=0 6 3 3 \ 2 /

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