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Hi,

We know that the series $\sum_{n=1}^\infty \frac{(-1)^{(n-1)}}{\sqrt{n}}$ is convergent and it is oscillating. and numerically it is almost 0.6048986434.

I want to know what is the exact limit of this series and how we can find that analytically.

Now if we let $A(t):=\sum_{n=1}^t\frac{(-1)^{(n-1)}}{\sqrt{n}}$, is there any simple formula without summation that is equal to $A(t)$.

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19  
I'm not sure what "exact limit" means to you - I mean, it's $(1-\sqrt{2})\zeta(1/2)$, but those are just some symbols. Regarding your question about $A(t)$: no. – David Hansen May 23 '11 at 22:42
8  
6048986434 is someone's phone number in Squamish, British Columbia, Canada, according to phonerecordfinder.com/number/604898 More to the point, your series shows up in Richard Johnsonbaugh, Summing an alternating series, Amer Math Monthly October 1979, 637-648. – Gerry Myerson May 24 '11 at 0:14
5  
Here's another way to ask what David Hansen is asking. What do you want to know about this series? In any application that I can think of, it suffices to know that the series converges conditionally in $\mathbb R$, diverges in all $\mathbb Q_p$, and is between $0$ and $1$. You know all this already. If you provide some motivation for why you're interested in this particular series, and some direction of the applications you're hoping for, perhaps you'll find useful answers. In the mean time, check out mathoverflow.net/faq and mathoverflow.net/howtoask . – Theo Johnson-Freyd May 24 '11 at 1:02

I am just stating David Hansen's comment as an answer, but with a bit of background. The way the question looks make it necessary to explain a few of the main components.

I suppose your question is:

Closed form of $$\sum_{n\geqslant 1}\frac{(-1)^{n-1}}{n^{1/2}}$$

This is a case of the Dirichlet Eta Function

$$\eta(s):=\sum_{n\geqslant1}\frac{(-1)^{n-1}}{n^{s}}=(1-2^{1-s})\zeta (s)$$

where $\zeta(s):=\sum_{n\geqslant 1} \frac{1}{n^s}$

Your result is $\eta(1/2)$ which is not known to have a closed form as $\zeta(1/2)$ doesn't have one known either. (91)

The second part to the comment is very telling, we do not know of a partial sum formula either.

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@Gerry Thank you for the edit – Small Margin May 25 at 23:29
1  
I think when you refer to $\zeta(1/2)$, you should not define $\zeta$ simply by that series. – Gerald Edgar May 26 at 1:56
    
@GeraldEdgar How might you define it? This is the main definition I am aware of. – Small Margin May 26 at 3:30
    
To define $\zeta(s)$ with real part $\le 1$, you have to add "analytic continuation" to the definition. The series does not converge for those values. – Gerald Edgar May 26 at 13:28

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