Summation of 1/i^2

rxt5972 · December 6th 2006, 12:23 AM

Hi guys,
The summation of i^2= n*(n+1)*(2n+1)/6
What is the summation od 1/i^2 ?
Thanks for your help.
rxt5972

**ThePerfectHacker** · December 6th 2006, 04:09 AM

Originally Posted by rxt5972

Hi guys,
The summation of i^2= n*(n+1)*(2n+1)/6
What is the summation od 1/i^2 ?
Thanks for your help.
rxt5972

I do not think there is one.

**Plato** · December 6th 2006, 02:24 PM

I saw your other double post. I agree with the response there.
You have been given an honest and correct answer,

However, I will give you a different answer.
The infinite series (sum) is: $\sum\limits_{k = 1}^\infty {\frac{1}{{k^2 }}} = \frac{{\pi ^2 }}{6}.$
Therefore, for large values of N we can use $\frac{{\pi ^2 }}{6}$ as an approximation of $\sum\limits_{k = 1}^N {\frac{1}{{k^2 }}}.$

**ThePerfectHacker** · December 6th 2006, 03:09 PM

Originally Posted by Plato

I saw your other double post. I agree with the response there.
You have been given an honest and correct answer,

However, I will give you a different answer.
The infinite series (sum) is: $\sum\limits_{k = 1}^\infty {\frac{1}{{k^2 }}} = \frac{{\pi ^2 }}{6}.$
Therefore, for large values of N we can use $\frac{{\pi ^2 }}{6}$ as an approximation of $\sum\limits_{k = 1}^N {\frac{1}{{k^2 }}}.$

Let me tell you why I do not think so.

Consider a similar example,
$\sum_{k=1}^n \frac{1}{k}$

And consider a Riemann integral over some positive closed interval on the function $f(x)=1/x$ . If we try to calculate this sum directly through the definition of the Riemann integral we will probably not be able to because the true values is the logarithmic function which is transendental.

**Plato** · December 6th 2006, 03:46 PM

Originally Posted by ThePerfectHacker

Let me tell you why I do not think so.
Consider a similar example,
$\sum_{k=1}^n \frac{1}{k}$
And consider a Riemann integral over some positive closed interval on the function $f(x)=1/x$ . If we try to calculate this sum directly through the definition of the Riemann integral we will probably not be able to because the true values is the logarithmic function which is transendental.

What in the world are you on about?
The harmonic series, $\sum_{k=1}^n \frac{1}{k}$ , is a very slowly divergent series.
The series $\sum_{k=1}^n \frac{1}{k^2}$ converges rather rapidly.
If fact for N=100 the estimate is correct in the first two decimal places.
For N=1000 the estimate is correct in the first four decimal places.

What is you point?

**ThePerfectHacker** · December 6th 2006, 04:10 PM

Originally Posted by Plato

What is you point?

I was working on a finite interval. So if you use a Riemann sum you need a way to add a finite harmonic series. Which will lead to a transendental function (the natural logarithm).

**Plato** · December 6th 2006, 04:32 PM

Originally Posted by ThePerfectHacker

I was working on a finite interval. So if you use a Riemann sum you need a way to add a finite harmonic series. Which will lead to a transendental function (the natural logarithm).

If N=1000000 the harmonic sum is less than 15.
But $\sum_{k=1}^n \frac{1}{k^2}$ is correct in first six decimal places.

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