TITLE: Please help me to prove the convergence of $\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}-\ln n$
QUESTION [6 upvotes]: Please help me to prove tha
 $$\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}-\ln n$$ converges and its limit is $\gamma \in [0,1]$. Then, using  $\gamma_{n}$ find the sum: $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$.

REPLY [2 votes]: The Mean Value Theorem says
$$
\frac{1}{k+1}<\log(k+1)-\log(k)<\frac{1}{k}
$$
Summing from $m$ to $n-1$ ($m < n$), we get
$$
\begin{align}
\sum_{k=m}^{n-1}\frac{1}{k+1}<\log(n)-\log(m)<\sum_{k=m}^{n-1}\frac{1}{k}
\end{align}
$$
Subtracting $\displaystyle{\sum_{k=m}^{n-1}\frac{1}{k+1}=\sum_{k=m+1}^n\frac{1}{k}=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^m\frac{1}{k}}$ from all parts, we get
$$
0<\left(\log(n)-\sum_{k=1}^n\frac{1}{k}\right)-\left(\log(m)-\sum_{k=1}^m\frac{1}{k}\right)<\frac{1}{m}-\frac{1}{n}
$$
This proves the existence of $\displaystyle{\lim_{n\to\infty}\left(\log(n)-\sum_{k=1}^n\frac{1}{k}\right)}$. If we set $m=1$ and subtract all sides from $1$, we get
$$
1>\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)>\frac{1}{n}
$$Thus, we have shown that $\displaystyle{\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)}$ exists, and that $0\le\gamma\le 1$.
Note that
$$
\begin{align}
\sum_{k=1}^{2n}(-1)^{k+1}\frac{1}{k}&=\sum_{k=1}^{2n}\frac{1}{k}-2\sum_{k=1}^n\frac{1}{2k}\\
&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\\
&=\left(\sum_{k=1}^{2n}\frac{1}{k}-\log(2n)\right)-\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)+\log(2)
\end{align}
$$
Taking the limit as $n\to\infty$, we get
$$
\sum_{k=1}^\infty(-1)^{k+1}\frac{1}{k}=\gamma-\gamma+\log(2)=\log(2)
$$