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

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SUMMARY

The infinite series 1/(k^1.5) converges to the value of ζ(3/2), where ζ represents the Riemann zeta function. The approximate value of this series is 2.6123753486854883433485675679240716305708006524, as confirmed by numerical summation and integration techniques. The series can be expressed using Laplace transforms and involves integrating a geometric series. However, finding a closed-form expression for ζ(3/2) remains a challenge, as closed forms for zeta values other than even integers are generally difficult to obtain.

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homology
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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,
 
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does it converge to something like 2.61216453017413...? that's what i got by adding the first 90 million terms.
 
Yeah I know, but how to show to what it converges exactly?
 
Here's something I've been trying. bear with me

Now

<br /> \frac{1}{(k^{1.5})}<br />

is the laplace transform of

<br /> \sqrt{t} \frac{2}{\sqrt{\pi}}<br />

So we can rewrite our series as

<br /> \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<br />

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

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

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

<br /> \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\sqrt{t}e^{-t}}{1-e^{-t}} dt<br />

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?
 
Last edited:
murshid_islam said:
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
 
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
 
homology said:
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? :smile:
 
homology said:
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.
 
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 height the second is <br /> \frac{1}{\sqrt{2}}<br />
The third is
<br /> \frac{1}{\sqrt{3}}<br />
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.
 

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