Infinite Sum Fallacy I was working on the infinite sum 
$$\sum_{x=1}^\infty \frac{1}{x(2x+1)}$$
and I used partial fractions to split up the fraction
$$\frac{1}{x(2x+1)}=\frac{1}{x}-\frac{2}{2x+1}$$
and then I wrote out the sum in expanded form:
$$1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+...$$
and then rearranged it a bit:
$$1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...$$
$$2-(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-...)$$
and since the sum inside of the parentheses is just the alternating harmonic series, which sums to $\ln 2$, I got
$$2-\ln 2$$
Which is wrong. What went wrong?
I notice that, in general, this kind of thing happens when I try to evaluate telescoping sums in the form
$$\sum_{x=1}^\infty f(x)-f(ax+b)$$
and I think something is happening when I rearrange it. Perhaps it has something to do the frequency of $f(ax+b)$ and that, when I spread it out to make it cancel out with other terms, I am "decreasing" how many of them there really are because I'm getting rid of the one to one correspondence between the $f(x)$ and $f(ax+b)$ terms?
I can't wrap my head around this. Please help!
 A: When you split the series in question via partial fractions you create an alternating series that happens to be conditionally convergent (it's not absolutely convergent), so rearranging 'a bit' is not allowed. What went wrong is exactly what you surmised: you've lost the correspondence between terms in the two 'halves' of your alternating series, and so you can achieve the wrong sum.
If you're careful with your partial sums, however, you can confirm the following:
$$
S_n := \sum_{k=1}^n \frac1{k(2k+1)}=\sum_{k=1}^n2\left(\frac1{2k}-\frac1{2k+1}\right)=2\left(1-A_{2n+1}\right),
$$
where $$A_n:=\sum_{k=1}^n\frac{(-1)^{k+1}}k$$
is the $n$th partial sum of the alternating harmonic series, and therefore $S_n$ converges to $2(1-\log 2)$.
A: By the Riemann rearrangement theorem, you have to be very careful when permuting terms of a conditionally but not absolutely convergent series. In your case it is probably simpler to notice that
$$ S=\sum_{n\geq 1}\frac{1}{n(2n+1)} = 2\sum_{n\geq 1}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{m\geq 2}\frac{(-1)^m}{m} $$
hence
$$ S = 2 \sum_{m\geq 2}(-1)^m \int_{0}^{1}x^{m-1}\,dx = 2\int_{0}^{1}\frac{x}{1+x}\,dx = \color{red}{2(1-\log 2)}.$$
A: This is one of the most astonishing things in mathematics: every conditionally convergent series (meaning: that converges but is not absolutely convergent, such as the alternate harmonic series you mentioned) can be conveniently rearranged to converge to any arbitrary real number, or just diverge.
This is the celebrated Riemann Series Theorem. You can search for a quick introduction to this theorem in Wikipedia, but if you want a proof of the theorem look for Fiktengolz's book on Fundamentals of Mathematical Analysis. I hope this could help you.
