Welcome to MathOverflow

Question

0

1

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 $A(t)$.

edited May 23 2011 at 22:25

asked May 23 2011 at 22:20

asd
73●5

12

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 2011 at 22:42

5

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 2011 at 0:14

4

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 2011 at 1:02

sum of the series $\sum_{n=1}^\infty \frac{(-1)^{(n-1)}}{\sqrt{n}}$?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.