Bound on Truncated Alternating Harmonic Series

Question

I'm trying to prove the following inequality

$$\sum _{n=1}^{2N}\frac{(-1)^n}{n}+\log (2)<\frac{1}{4N+1}.$$

Most of the traditional inequalities I've seen for harmonic numbers aren't tight enough to prove this, unless I'm doing something wrong.

Answer 1

As the taylor series of $\ln(1+x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n-1}x^n}{n}$, we have

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

then \begin{align} \sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n} + \ln(2) &= \sum_{n=2N+1}^{+\infty} \frac{(-1)^{n-1}}{n} \\ &= \sum_{k=N}^{+\infty} \left( \frac{(-1)^{2k}}{2k+1}+\frac{(-1)^{2k+1}}{2k+2} \right) \\ &= \sum_{k=N}^{+\infty} \left( \frac{1}{2k+1}-\frac{1}{2k+2} \right) \\ &= \sum_{k=N}^{+\infty} \frac{1}{(2k+1)(2k+2)} \\ &= \sum_{k=N}^{+\infty} \frac{4}{(4k+2)(4k+4)} \\ &< \sum_{k=N}^{+\infty} \frac{4}{(4k+1)(4k+5)} =\sum_{k=N}^{+\infty} \left( \frac{1}{4k+1}-\frac{1}{4(k+1)+1} \right) = \frac{1}{4N+1}\\ \end{align}