# Problem concerning the sequence $s_n = 1 + 1/2 +\cdots+ 1/n - \log n$

The question is :

Prove that the sequence $\{s_n\}$ where $s_n = 1 + 1/2 +\cdots+ 1/n - \log n$ is convergent. Hence find $\lim_{n \to \infty} \left(1 - 1/2 + 1/3 - ... - 1/2n\right)$.

I have proved that the sequence $\{s_n\}$ is monotonic decreasing and bounded below by $0$ and hence it is convergent.So,first part of the question is done.Now if $t_n = 1 - 1/2 + 1/3 - \cdots - 1/2n$ then I have independently (without taking help of the convergence of the sequence $\{s_n\}$) proved that the sequence $\{t_n\}$ is monotonic increasing and bounded above by $1$ and hence it is convergent.Now how can show it with the help of the first part.Also how can evaluate the limit of the later though I know that it converges to $\log2$. Isn't it? Please help me. Thank you in advance.

One may write, as $n \to \infty$, \begin{align} &1 - \frac12 + \frac13 - \cdots+ \frac1{2n-1}- \frac1{2n} \\\\&=\left(1 + \frac12 + \frac13 + \dots + \frac1{2n}\right)-2\left( \frac12 + \frac14 + \frac16 + \dots + \frac1{2n}\right) \\\\&=\left(1 + \frac12 + \frac13 + \dots + \frac1{2n}\right)-\left( 1 + \frac 12 + \frac 13 + \dots + \frac 1n\right) \\\\&=\left(1 + \frac12 + \frac13 + \dots + \frac1{2n}-\ln(2n)\right)-\left( 1 + \frac 12 + \frac 13 + \dots + \frac 1n-\ln n\right)+\ln 2 \\\\&=s_{2n}-s_n+\ln 2 \end{align} then use what you have proved in the first part, $$\lim_{n \to \infty}s_{2n}=\lim_{n \to \infty}s_n<\infty$$to conclude that, as $n \to \infty$,
$$1 - \frac12 + \frac13 - \cdots+ \frac1{2n-1}- \frac1{2n} \to \ln 2.$$
Since $n=\prod_{k=1}^{n-1}\left(\frac{k+1}{k}\right)$ we have: $$H_n-\log(n+1) = \sum_{k=1}^{n}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right) \tag{1}$$ where $f(x)=x-\log(1+x)$ is bounded between $0$ and $\frac{x^2}{2}$ on the interval $[0,1]$.
It follows that the sequence given by $a_n = H_n-\log(n+1)$ is increasing and bounded by $\frac{\pi^2}{12}$.
Now $H_n=\log(n)+O(1)$ also gives: $$1-\frac{1}{2}+\ldots-\frac{1}{2n} = H_{2n}-H_n = \color{red}{\log 2}+o(1).\tag{2}$$