How Do You Calculate the Sum of the Series 1/(k^1.5)?

[homology]

hello,

I'm working on a little puzzle and part of it requires summing the infinite series 1/(k^1.5) which clearly converges, but I've never been very good at actually finding what a series converges to. Could you give me a good swift kick in the head. Just a hint will do.

Thanks,

 

[murshid_islam]

does it converge to something like 2.61216453017413...? that's what i got by adding the first 90 million terms.

 

[homology]

Yeah I know, but how to show to what it converges exactly?

 

[homology]

Here's something I've been trying. bear with me

Now

$\frac{1}{(k^{1.5})}$

is the laplace transform of

$\sqrt{t} \frac{2}{\sqrt{\pi}}$

So we can rewrite our series as

$\sum_{k=1}^{\infty} \frac{1}{k^{1.5}} = \sum_{k=1}^{\infty} \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \sqrt{t} e^{-kt} dt$

The integral is pretty nice being positive and decreasing so there's probably a nice theorem out there saying we can swap sum and integral, which gives us

$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \sqrt{t} \sum_{k=1}^{\infty} e^{-kt} dt$

The sum is a geometric one, or pretty close, and we get the following

$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\sqrt{t}e^{-t}}{1-e^{-t}} dt$

Now I'm a big baby and can't figure out how to integrate this. I shove it in mathematica (and replace infinity with a big number) and I get the same (approximately) as I do when evaluate the truncated series. Which is about 2.61238. Now I'm sure there's a nice way to evaluate this integral and get an exact answer. Any takers?

 

[CRGreathouse]

does it converge to something like 2.61216453017413...? that's what i got by adding the first 90 million terms.

Well, you have the first two decimal places correct. It's 2.6123753486854883433485675679240716305708006524...

It's just the zeta function, which is easy to calculate.
http://mathworld.wolfram.com/RiemannZetaFunction.html

 

[homology]

Well I looked into the link and I can see lots of information on the zeta function evaluated at integer values but nothing about fractional powers like 3/2. Could you outline how to get an exact expression for something like this?

Thanks

Kevin

 

[Data]

Well I looked into the link and I can see lots of information on the zeta function evaluated at integer values but nothing about fractional powers like 3/2. Could you outline how to get an exact expression for something like this?

Thanks

Kevin

it converges to $\zeta (\frac{3}{2})$. What makes you think there's a nicer way to write it down?

 

[CRGreathouse]

Well I looked into the link and I can see lots of information on the zeta function evaluated at integer values but nothing about fractional powers like 3/2. Could you outline how to get an exact expression for something like this?

There are several mentions of fractional values, just none for 3/2. I doubt there's a closed form expression; finding closed-form expressions for zeta values other than even integers is hard in general.

 

[homology]

I only thought it converged to a nicer expression because its part of a puzzle I was working on. You are given a gift that is unusually wrapped. It is shaped like a stack of cubes, the first one is 1 foot hight the second is $\frac{1}{\sqrt{2}}$
The third is
$\frac{1}{\sqrt{3}}$
and so on. So the height is clearly infinite. The problem also asks about the surface area, which is infinites and the volume which gives us the series I started with, which is finite. Given that it was a puzzle, I sort of supposed that it would come out nicer. Thanks for all the input.