Limit of definite sum equals ln(2)

  • Thread starter Berrius
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  • #1
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Homework Statement


As part of a problem I have to show that [tex]lim_{n\to\infty}\sum_{i=\frac{n}{2}}^{n}\frac{1}{i}=ln(2)[/tex]


Homework Equations


Taylor expansion of ln(2): [tex]\sum_{i=1}^{\infty}\frac{(-1)^{k+1}}{k}[/tex]


The Attempt at a Solution


ln(2) can be written as: [tex]ln(2) = \sum_{i=1}^{\frac{n}{2}}\frac{(-1)^{k+1}}{k} + \sum_{i=\frac{n}{2}+1}^{n}\frac{(-1)^{k+1}}{k} + \sum_{i=n+1}^{\infty}\frac{(-1)^{k+1}}{k}[/tex]

Where the middle term looks alot like the sum i need. However I need some way to get rid of the alternating sign change, but i don't see how.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


As part of a problem I have to show that [tex]lim_{n\to\infty}\sum_{i=\frac{n}{2}}^{n}\frac{1}{i}=ln(2)[/tex]


Homework Equations


Taylor expansion of ln(2): [tex]\sum_{i=1}^{\infty}\frac{(-1)^{k+1}}{k}[/tex]


The Attempt at a Solution


ln(2) can be written as: [tex]ln(2) = \sum_{i=1}^{\frac{n}{2}}\frac{(-1)^{k+1}}{k} + \sum_{i=\frac{n}{2}+1}^{n}\frac{(-1)^{k+1}}{k} + \sum_{i=n+1}^{\infty}\frac{(-1)^{k+1}}{k}[/tex]

Where the middle term looks alot like the sum i need. However I need some way to get rid of the alternating sign change, but i don't see how.
You can use the exact result that
[tex] \sum_{i=1}^n \frac{1}{i} = \Psi(n+1) + \gamma,[/tex]
where ##\gamma## is Euler's constant and
[tex] \Psi(x) = \frac{\Gamma^{\, \prime}(x)}{\Gamma(x)}[/tex]
is the so-called "di-gamma" function. This function has numerous published properties, which should allow you to derive the answer you want.
 
  • #3
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Thanx. But I think this approach is beyond the scope of my course. I was specifically told to look at the taylor expansion of ln(2). Is there a way to do it that way?
 

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