Estimate the sum of alternating harmonic series between $7/12$ and $47/60$ How can I prove that: 
$$\frac{7}{12} < \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} < \frac{47}{60}$$ 
? I don't even know how to start solving this... 
 A: First note that the series converges using Leibniz Test.
Next, denote by $S_N$ the partial sum $\sum_{n=1}^N\frac{(-1)^{n-1}}{n}$.  Then, we must have $$S_{2N}<\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}<S_{2N+1}$$
Finally, we see that $\sum_{n=1}^4 \frac{(-1)^{n+1}}{n}=\frac{7}{12}$ and inasmuch as the next term is positive, the value of the series must exceed $7/12$.  Similarly, we see that $\sum_{n=1}^5 \frac{(-1)^{n+1}}{n}=\frac{47}{60}$ and inasmuch as the next term is negative, the value of the series must be less than $47/60$.
A: If $s_{n}:=\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}}{k}$ then $\left(s_{2n-1}\right)_{n=1,2,\dots}$
is monotonically decreasing and $\left(s_{2n}\right)_{n=1,2,\dots}$
is monotonically increasing. 
This with $s_{2n}<s_{2n-1}$. 
So if you can find some $n$ with $\frac{7}{12}<s_{2n}<s_{2n-1}<\frac{47}{60}$
then you are ready.
A: Hint your sum is integration of $\int (1+x)^{-1}$ after integration put $x=1$ . So series converges to $\ln(2)$
A: The manipulation
$$\sum_{n\geq 0}\frac{(-1)^{n}}{n+1} = \int_{0}^{1}\sum_{n\geq 0}(-1)^n x^n\,dx = \int_{0}^{1}\frac{dx}{x+1} = \int_{0}^{1}\frac{dx}{2-x}=\sum_{n\geq 0}\frac{1}{(n+1)\,2^{n+1}}$$
gives a substantial convergence boost, and summation by parts gives a second boost:
$$ \sum_{n\geq 0}\frac{1}{(n+1) 2^{n+1}} = 1-\sum_{n\geq 0}\frac{1}{(n+1)(n+2)2^{n+1}}\tag{1} $$
By evaluating $\sum_{n=0}^{4}\frac{1}{(n+1)2^{n+1}}$ and $1-\sum_{n=0}^{3}\frac{1}{(n+1)(n+2)2^{n+1}}$ we get the improved inequality:

$$ \color{red}{\frac{661}{960}}\leq \sum_{n\geq 0}\frac{(-1)^n}{n+1}\leq\color{red}{\frac{667}{960}}.\tag{2}$$

