Limit of definite sum equals ln(2)

[Berrius]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

ln(2) can be written as: 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}

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

ln(2) can be written as: 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}

Where the middle term looks a lot 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
\sum_{i=1}^n \frac{1}{i} = \Psi(n+1) + \gamma,
where $\gamma$ is Euler's constant and
\Psi(x) = \frac{\Gamma^{\, \prime}(x)}{\Gamma(x)}
is the so-called "di-gamma" function. This function has numerous published properties, which should allow you to derive the answer you want.

 

[Berrius]

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?