Limit of ratio involving harmonic numbers and ln 2

[nonaa]

Homework Statement

$$\lim_{n\to\infty}\frac{\ln 2 - \sum_{k=1}^{n}\frac{1}{k+n}}{\ln 2 - \sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}}=?$$

Homework Equations

$$H(2n)-H(n)= \sum_{k=1}^{n}\frac{1}{k+n}$$

The Attempt at a Solution

I tried to use that $$\ln 2 = \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}{k}$$, but with no success. Can you help me, please?

 

[nonaa]

Any ideas?

 

[snipez90]

The sum term in the numerator approaches ln(2) as n goes to infinity since the sum can be manipulated into resembling a Riemann sum (via a regular n equal subdivisions partition of [0,1]). Also what exactly is H?

 

[nonaa]

yes, I already found that, but how does it help us?
H - harmonic series: $$H(n)=\sum_{k=1}^{n}\frac{1}{k}$$; It's not part of the problem, just thought it can be useful...

 

[Dick]

You can write complicated expressions for the numerator and denominator in terms of the digamma function. Which means you can apply l'Hopital. What can do with that depends on how much you know about the digamma functions. I don't know much. Just playing with the numbers seems to show the absolute value of that ratio approaches a limit. But since the numerator is an alternating series approximation the sign keeps flipping.

 

[nonaa]

Sadly, I know nothing about the digamma function :(