Re: Closed form for series

David W. Cantrell wrote:
"I.N. Galidakis" <morpheus@xxxxxxxxxxxx> wrote:
Does anyone know a closed form for:

sum((-1)^n*log(n),n=2..oo)?

Unsigned oo ?

In what sense do you think the series might be said to converge?

It might help you to know that

sum((-1)^n*log(n),n=2..N) =

1/2*(log(pi/2) + (-1)^N*log((2*Gamma((2 + N)/2)^2)/Gamma((1 +
N)/2)^2))

David

Thanks to both, David and Rob. It obviously doesn't converge (using conventional
summation), but Maple 9 reports the value:

0.2257913526, which is consistent with the following:

This series arose as I was trying to prove the identity:

Pi/2=product((n/(n+1))^((-1)^n),n=1..infinity), (1) which is the series:

(1/2)^(-1)*(2/3)*(3/4)^(-1)*(4/5)*(5/6)^(-1)*...

This is the same as the Wallis Product series:

(2/1)*(2/3)*(4/3)*(4/5)*....

http://en.wikipedia.org/wiki/Wallis_product

Hence:

Pi/2=product((n/(n+1))^((-1)^n),n=1..infinity).

Taking logs on both sides of (1), we get:

log(Pi/2)=sum((-1)^n*log(n/(n+1)),n=1..infinity)
=sum((-1)^n*[log(n)-log(n+1)],n=1..infinity)
=-log(1)
+log(2)+log(2)
-log(3)-log(3)
+....
=2*sum((-1)^n*log(n),n=2..infinity) =>

1/2*log(Pi/2)=sum((-1)^n*log(n),n=2..infinity)

The value that Maple reports, satisfies the above identity. Note also that the
term on the left is the same as the first term in David's expression, provided
the rest goes to 0: 1/2*(log(pi/2) + (-1)^N*log((2*Gamma((2 + N)/2)^2)/Gamma((1
+ N)/2)^2)).

But it does not. Doing a plot of the David's expression for N \in {1..1000},
shows that the second part of the expression becomes undefined as N->oo.

Is my taking the log of an infinite product silly? I've seen it done in proving
that some infinite products converge, by reducing them to infinite series.

So, what the heck is going on here?
--
I.N. Galidakis

.

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