## An alternating sum

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### An alternating sum

Hello, I have recently encountered the sum ln(1)/1 - ln(2)/2 + ln(3)/3 - ln(4)/4 + ln(5)/5 - ln(6)/6 ..... = sum n=1 to infinity -(-1)^n*ln(n)/n
It is clear it converges, but does anyone know the answer and have a proof? Thank you!

cyanyoshi
Posts: 418
Joined: Thu Sep 23, 2010 3:30 am UTC

### Re: An alternating sum

Just because a series converges does not mean that there is necessarily a simple, pretty expression that it simplifies to. After a bit of hunting around on Wikipedia, this series may equal d/ds Lis(-1) evaluated at s = 1, where Lis(z) is the polylogarithm function. Whether this is in any way more elegant than the original expression is up for debate, though. There are probably at least a few other ways to write this using existing functions and operators.

Spoiler:
Lis(z) ≡ ∑k=1 zk/ks
Lis(-1) = ∑k=1 (-1)k/ks
d/ds Lis(-1) = ∑k=1 d/ds (-1)k/ks (not 100% sure if differentiating term-by-term is ok)
d/ds Lis(-1) = ∑k=1 d/ds (-1)k k-s
d/ds Lis(-1) = ∑k=1 -(-1)k ln(k) k-s
d/ds Lis(-1)|s=1 = ∑k=1 -(-1)k ln(k) / k
Last edited by cyanyoshi on Sat Jul 21, 2012 7:10 am UTC, edited 1 time in total.

Posts: 36
Joined: Mon May 28, 2012 7:11 pm UTC

### Re: An alternating sum

Thank you very much, while that probably won't help me find a closed form that I would like, it is very interesting, as Lis(-1) is related to the Dirichlet eta function (and is in fact it's opposite) , and right above references on this page: http://en.wikipedia.org/wiki/Dirichlet_eta_function is a formula for the derivative that I want... Which unfortunately is undefined at s=1 , and while the limit should exist I have no idea on how to evaluate it... I believe this sum has a closed form with natural logarithms and the Euler-Mascheroni constant because of some integrals I've been fooling around with, but I'm not quite sure yet. Thank you for your help!

PM 2Ring
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Location: Sydney, Australia

### Re: An alternating sum

If it's any help, I get a value of -0.159868(9), using brute-force summation to 10000 terms, followed by Richardson approximation.

To 25000 terms:
-0.1598688994744

although I suspect that the last 4 or 5 digits are garbage.

Posts: 36
Joined: Mon May 28, 2012 7:11 pm UTC

### Re: An alternating sum

Thank you, although I probably do not need too many digits, just enough to make me feel confident with my answer, once I get it. Wolfram-Alpha spits out hundreds of digits, although I'm sure it messes up quite quickly, who knows. All I know is that I have plenty of digits, thanks for the support! I know I will solve it, just one more crazy integral to go...

eta oin shrdlu
Posts: 451
Joined: Sat Jan 19, 2008 4:25 am UTC

### Re: An alternating sum

Heptadecagon wrote:I believe this sum has a closed form with natural logarithms and the Euler-Mascheroni constant because of some integrals I've been fooling around with, but I'm not quite sure yet. Thank you for your help!
You are right. Here's one approach:
Spoiler:
Consider the finite sum (to avoid convergence issues) and split it into two nonalternating series:

2Nn=1 (-1)n-1 ln n/n
= 2Nn=1 ln n/n - 2 Nn=1 ln(2n)/(2n)
= 2Nn=N+1 ln n/n - ln 2 Nn=1 1/n

It's easy to see that the second series is roughly ln 2 ln N, and
limN→∞ ln 2 Nn=1 1/n - ln 2 ln N = γ ln 2 .

You can turn the first series into Riemann sums,
2Nn=N+1 (ln N + ln(n/N))/n
= ln N 2Nn=N+1 1/(n/N) 1/N + 2Nn=N+1 ln(n/N)/(n/N) 1/N .

The first of these series approaches
ln N 12 dx/x = ln N ln 2
which cancels the ln 2 ln N divergence in the limit above.

The second of these series approaches
12 ln x/x dx = (ln 2)2/2 .

You can be more rigorous about taking the N→∞ limit to keep everything convergent; this gives the closed form
(ln 2)2/2 - γ ln 2 .
This sum was recently posed at Les Reid's site; I don't know its earlier history.

Posts: 36
Joined: Mon May 28, 2012 7:11 pm UTC

### Re: An alternating sum

Cool, thanks for the help, it's interesting to see such different methods. I've already seen the -ln(2)*γ term, but not the other quite yet. I understand the proof you posted except to
''The second of these series approaches
∫12 ln x/x dx = (ln 2)2/2 . '' , but I have just woke up and haven't got out any paper yet. Thanks for all the help, and even if someone has another method of solving it it would be interesting to see, and it would also be neat to know the history of this sum. Thanks!

PM 2Ring
Posts: 3715
Joined: Mon Jan 26, 2009 3:19 pm UTC
Location: Sydney, Australia

### Re: An alternating sum

eta oin shrdlu wrote:
Spoiler:
this gives the closed form
(ln 2)2/2 - γ ln 2 .

Ah. That's approximately equal to
Spoiler:
-0.
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