Please help me to prove the convergence of $\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}-\ln n$ 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: $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$.
 A: The following question addresses the convergence, so assume $\gamma_n$ converges and denote the limit as $\gamma$.
Now consider  $\sum_{n=1}^{2m} (-1)^{n+1} \frac{1}{n}$ and split it into even and odd terms
$\sum_{n=1}^m \frac{1}{2n-1} - \sum_{n=1}^m \frac{1}{2n}$. Then complete the sum over odd integers to sum over consequtive integers: $\sum_{n=1}^{2m} \frac{1}{n} - 2 \sum_{n=1}^m \frac{1}{2n}$. 
Then subtract logarithms to form $\gamma_{2m} - 2 \gamma_m + \log(2)$, like so 
$$ ( \sum_{n=1}^{2m} \frac{1}{n} - \log(2m)) - ( \sum_{n=1}^m \frac{1}{n} - \log(m)) + \log(2).$$
In the limit $m \to \infty$ it becomes $ \gamma - \gamma + \log(2) = \log(2)$.
A: You can use the mean value theorem to prove first that
$$\frac{1}{n+1}<\ln(n+1)-\ln n<\frac{1}{n}$$
then, $\ln(n+1)<1+\ldots+\frac{1}{n}<1+\ln n$, and finally
$$\ln(n+1)-\ln n<\gamma_n<1$$
which implies the convergence of $\gamma_n$ and its limit $\gamma\in[0,1]$.
For the second part, I am not sure if you can use $\gamma_n$ to find the sum $\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}$! 
I suggest to use the Taylor-Young formula for $x\mapsto\ln(x+1)$ on $[0,1]$ to prove that 
$$\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}=\ln2.$$
A: 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
$$
\sum_{k=1}^\infty(-1)^{k+1}\frac{1}{k}=\gamma-\gamma+\log(2)=\log(2)
$$
A: Do we have a "proof without words" question here?  I don't find it.  There is a lovely diagram found in certain calculus texts that does this for us.  Maybe I will animate it, for fun.
