A sum with irrational limits

Aug 2018
128
7
România
Good evening to all,

Calculate the partial sums \(\displaystyle \sum_{k=\sqrt{2}}^{\sqrt{n}} \frac{1}{k(k+1)}\) , where \(\displaystyle n\in \mathbb N^*\).

All the best,

Integrator
 
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romsek

Math Team
Sep 2015
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USA
That notation is isn't standard. Are we to assume the index variable takes values $\sqrt{2}, \sqrt{3}, \dots, \sqrt{n}$ ?
 
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romsek

Math Team
Sep 2015
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perhaps a better way to write it would be

$\sum \limits_{k=2}^n\dfrac{1}{\sqrt{k}(\sqrt{k}+1)}$

I don't think this is going to have a closed form. Usually you can apply partial fractions and the series of this type will seen be to telescoping but here

$\sqrt{k+1} \neq \sqrt{k}+1$

so this doesn't occur.

Mathematica can't make anything out of it.
 
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Dec 2015
928
122
Earth
Try to express the general term as \(\displaystyle \dfrac{1}{\sqrt{x}(1+\sqrt{x})}=\dfrac{\sqrt{x}}{x(1-x)}-\dfrac{1}{1-x}\).
 

topsquark

Math Team
May 2013
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Try to express the general term as \(\displaystyle \dfrac{1}{\sqrt{x}(1+\sqrt{x})}=\dfrac{\sqrt{x}}{x(1-x)}-\dfrac{1}{1-x}\).
Okay, I'm stumped. How does this help?

-Dan
 
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Aug 2018
128
7
România
perhaps a better way to write it would be

$\sum \limits_{k=2}^n\dfrac{1}{\sqrt{k}(\sqrt{k}+1)}$

I don't think this is going to have a closed form. Usually you can apply partial fractions and the series of this type will seen be to telescoping but here

$\sqrt{k+1} \neq \sqrt{k}+1$

so this doesn't occur.

Mathematica can't make anything out of it.
Hello,

Are you saying that the sum \(\displaystyle \sum_{k=\sqrt{2}}^{\sqrt{n}} \frac{1}{k(k+1)} \) , where \(\displaystyle n\in \mathbb N^*\) cannot be calculated?

All the best,

Integrator
 
Last edited:

romsek

Math Team
Sep 2015
2,751
1,531
USA
Hello,

Are you saying that the sum \(\displaystyle \sum_{k=\sqrt{2}}^{\sqrt{n}} \frac{1}{k(k+1)} \) , where \(\displaystyle n\in \mathbb N^*\) cannot be calculated?

All the best,

Integrator
for any given $n$ sure it can be calculated but leaving $n$ as a parameter I don't think the sum has any closed form.
 
Dec 2015
928
122
Earth
To make it clear : \(\displaystyle s_n = \dfrac{1}{\sqrt{2}(1+\sqrt{2})}+\dfrac{1}{\sqrt{3}(1+\sqrt{3})}+\dotsc + \dfrac{1}{\sqrt{n}(1+\sqrt{n})}\).

or \(\displaystyle s_n= [1+\dfrac{1}{2}+\dfrac{1}{3}+\dotsc + \dfrac{1}{n-1} ] - [\dfrac{\sqrt{2}}{2^2 -2}+\dfrac{\sqrt{3}}{3^2 -3}+\dotsc + \dfrac{\sqrt{n}}{n^2 -n} ]\).
 
Last edited:
Aug 2018
128
7
România
To make it clear : \(\displaystyle s_n = \dfrac{1}{\sqrt{2}(1+\sqrt{2})}+\dfrac{1}{\sqrt{3}(1+\sqrt{3})}+\dotsc + \dfrac{1}{\sqrt{n}(1+\sqrt{n})}\).

or \(\displaystyle s_n= [1+\dfrac{1}{2}+\dfrac{1}{3}+\dotsc + \dfrac{1}{n-1} ] - [\dfrac{\sqrt{2}}{2^2 -2}+\dfrac{\sqrt{3}}{3^2 -3}+\dotsc + \dfrac{\sqrt{n}}{n^2 -n} ]\).
Hello,

I do not understand!
For example , others say that for \(\displaystyle 2\leq n\leq 5\) we get \(\displaystyle \sum_{k=\sqrt{2}}^{\sqrt{n}} \frac{1}{k(k+1)}=\frac{\sqrt{2}-1}{\sqrt{2}}\) , where \(\displaystyle n\in \mathbb N^*\).Is this result correct?Thank you very much!

All the best,

Integrator