Find constants s.t. the following expression holds for all n.

[phreak]

Homework Statement

Find constants $$c_1,c_2$$ (independent of n) such that the following holds for all $$n\in \mathbb{N}$$:

$$\left| \sum^{2n}_{k=n+1} \frac{1}{k} - \log 2 - \frac{c_1}{n} \right| \le \frac{c_2}{n^2}.$$

Homework Equations

$$\log(2) = \sum^{\infty}_{k=1} (-1)^{k+1}\left( \frac{1}{k} \right).$$

The Attempt at a Solution

Well, the obvious method would be to just input the series for $$\log(2)$$ and try to find something from that. I tried but found virtually nothing that I could work with. Can anyone give me any hints as to how to proceed?

Edit: Okay, I think I have just confirmed that

$$\sum^{2n}_{k=n+1} \frac{1}{k} - \log 2 = -\sum^{\infty}_{k=2n+1} (-1)^{k+1} \left( \frac{1}{k} \right).$$

 

[lippyka]

If that is correct, then it would seem that we could use the fact that \left| \sum^{\infty}_{k=2n+1} (-1)^{k+1} \left( \frac{1}{k} \right) \right| \le \sum^{\infty}_{k=2n+1} \left( \frac{1}{k} \right) = \frac{1}{2n}.Thus, we can set c_2 = 1/2, and thus we have \left| \sum^{2n}_{k=n+1} \frac{1}{k} - \log 2 - \frac{c_1}{n} \right| \le \frac{1/2}{n^2}.Is this correct? Edit 2: Now I'm thinking that the answer should be c_2 = 1, since we must also account for the \frac{c_1}{n} term. Is this correct?