Find the value of $\sum_{n=1}^{\infty} \frac{2}{n}-\frac{4}{2n+1}$ Find the value of $$S=\sum_{n=1}^{\infty}\left(\frac{2}{n}-\frac{4}{2n+1}\right)$$  
My Try:we have
$$S=2\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{2n+1}\right)$$  
$$S=2\left(1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+\cdots\right)$$ so
$$S=2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots\right)$$  But we know
$$\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$  So
$$S=2(2-\ln 2)$$
Is this correct?
 A: Note that this is not valid by the Riemann series theorem, which shows you cannot group terms like that.  In particular, the terms you are grouping tend to be farther and farther from each other, meaning you are "pulling" terms faster than others, and this results in the value of the series changing, since $\sum\frac1n$ does not converge.  Indeed, note that:
$$S=4\sum_{n=1}^\infty\frac1{2n}-\frac1{2n+1}=4\sum_{n=2}^\infty\frac{(-1)^n}{n}=4(1-\ln(2))$$
Which is different from your result.  Indeed, if we consider the partial sums correctly, this is the correct result.

A more explicit example of the Riemann series theorem:
$$0=\sum_{n=1}^\infty\frac1n-\frac1n$$
Note that $\frac1n$ is either $\frac1{2n}$ or $\frac1{2n-1}$, hence, we group more positive terms together:
$$\begin{align}0&\stackrel?=\sum_{n=1}^\infty\frac1{2n-1}+\frac1{2n}-\frac1n\\&=\sum_{n=1}^\infty\frac1{2n-1}-\frac1{2n}\end{align}$$
But now notice that $\frac1{2n-1}-\frac1{2n}>0$ for all $n$, hence,
$$0\stackrel?>0$$
Which should be intuitive.  Since we added up the positive terms faster, the resulting series became larger.

If you can't see the manipulation step, here it is written out:
$$\begin{align}0&=\color{#4488ee}{\frac11}-\frac11+\color{#4488ee}{\frac12}-\frac12+\color{#44ee88}{\frac13}-\frac13+\color{#44ee88}{\frac14}-\frac14+\color{orange}{\frac15}-\frac15+\color{orange}{\frac16}-\frac16+\dots\\&\stackrel?=\color{#4488ee}{\frac11+\frac12}-\frac11+\color{#44ee88}{\frac13+\frac14}-\frac12+\color{orange}{\frac15+\frac16}-\frac13+\frac17+\frac18-\frac14+\dots\\&\stackrel?=\frac11+\left(\frac12-\frac11\right)+\frac13+\left(\frac14-\frac12\right)+\frac15+\left(\frac16-\frac13\right)+\frac17+\left(\frac18-\frac14\right)+\dots\\&\stackrel?=\frac11-\frac12+\frac13-\frac14+\frac16-\frac14+\dots\\&\stackrel?=\ln(2)\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
S & \equiv \sum_{n = 1}^{\infty}\pars{{2 \over n} - {4 \over 2n + 1}} =
\sum_{n = 1}^{\infty}\pars{{4 \over 2n} - {4 \over 2n + 1}} =
4\sum_{n = 2}^{\infty}{\pars{-1}^{n} \over n} =
4\bracks{1 + \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}}
\\[5mm] & =
4\pars{\vphantom{\LARGE A}1 +
\braces{\vphantom{\Large A}-\ln\pars{\vphantom{\large A}1 - \bracks{-1}}}} =
\bbx{4\bracks{\vphantom{\large A}1 - \ln\pars{2}}}
\end{align}

Indeed, it's quite close to $\texttt{@Simply Beautiful Art}$ fine answer.

A: We can use the digamma function $\psi$. It is known that
$$\psi(z+1)=-\gamma+\sum_{n=1}^{\infty}\dfrac{z}{n(n+z)},\ \ \forall z \in \mathbb{C} \setminus \{-1,-2,...\}.$$
Thus for $z=\frac{1}{2}$ we obtain
$$\psi\left(\dfrac{3}{2}\right)=-\gamma+\sum_{n=1}^{\infty}\dfrac{1}{n(2n+1)}=-\gamma+\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{2}{(2n+1)}\right)=-\gamma+\dfrac{S}{2}.$$
From the other hand, for the half-integer values of $\psi$ we have 
$$\psi\left(n+\dfrac{1}{2}\right)=-\gamma-2\ln 2+\sum_{k=1}^n\dfrac{2}{2k-1},$$
from where, for $n=1$, we conclude that
$$\psi\left(\dfrac{3}{2}\right)=-\gamma-2\ln 2+2.$$
Finally $-\gamma+\dfrac{S}{2}=-\gamma-2\ln 2+2 \Rightarrow S=4(1-\ln 2),$
as Simply Beautiful Art mentioned.
Note: S is a real number and it can be justified via comparison with the series of the Basel Problem.
