Find the sum $1+\frac13+\frac15-\frac12-\frac14+$… (to clarify, this is a rearrangement of the alternating harmonic series with 3 positive terms followed by 2 negative terms repeatedly)
I know by the Riemann theorem that rearranging the elements of a conditionally convergent series can give any real sum.
However, I don't know of any power series that can express this sum, so I don't know how to find it.
Best Answer
Let $t_n=1+\frac13+\frac15-\frac12-\frac14\cdots\text{upto n terms}$
And $\gamma_n=(1+\frac12+\frac13\cdots\frac1n)-\log n$
$\lim_{n\to\infty}\gamma_n=\gamma$ this is the Euler Mascheroni constant
So $$\begin{align}t_{5n}=&(1+\frac13+\frac15-\frac12-\frac14)+(\frac17+\frac19+\frac1{11}-\frac16-\frac18)\cdots+(\frac1{6n-5}+\frac1{6n-3}+\frac1{6n-1}-\frac1{4n-2}-\frac1{4n})\\&=(1+\frac13+\frac15+\frac17\cdots+\frac1{6n-1})-\frac12(1+\frac12+\frac13\cdots\frac1{2n})\\&=(1+\frac12+\frac13\cdots+\frac1{6n})-(\frac12+\frac14\cdots\frac1{6n})-\frac12(1+\frac12\cdots+\frac1{2n})\\&=(\gamma_{6n}+\log 6n)-\frac12(\gamma_{3n}+\log 3n)-\frac12(\gamma_{2n}+\log2n)\\&=\frac12\log6+\gamma_{6n}-\frac12(\gamma_{2n}+\gamma_{3n})\end{align}$$
As n tends to $\infty$
$$t_\infty=\frac12\log6+\gamma-\frac12(\gamma+\gamma)=\frac12\log6$$