Stupid question about $1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$

Question

I have

$$1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$$

Partial sum $S_{3n}$ of the above is:

$$(1 - \frac{1}{2}-\frac{1}{4})+(\frac{1}{3}- \frac{1}{6}-\frac{1}{8})+(\frac{1}{5}-\dots$$

But what is $S_{3n-1}$ and $S_{3n-2}$ ?

$S_{3n-1}=S_{3n}-a_{3n}$, though I don't recognize the pattern that gives us $a_{3n }$ — Hagen von Eitzen, May 31 '13 at 14:45
We could stand to see a few more terms... maybe up to the twelfth? — J. M. ain't a mathematician, May 31 '13 at 14:47
@J.M.: I'm fairly certain that it's a summation of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}n$, taking two even denominators after each odd denominator. — Glen O, May 31 '13 at 14:51
@Glen, I did recognize the rearranged alternating harmonic series, but I really was hoping the OP had explicitly stated "two even denominators after each odd denominator" or something to that effect. — J. M. ain't a mathematician, May 31 '13 at 14:52

score 1 · Accepted Answer · answered May 31 '13 at 15:02

By guessing the pattern hidden in the ellipsis, it seems that you consider the series $\sum_{k=1}^\infty a_k $ where $a_{3n-2}=\frac1{2n-1}$, $a_{3n-1}=-\frac1{4n-2}$, $a_{3n}=-\frac1{4n}$ for $n\ge 1$. Thus $a_{3n-2}+a_{3n-1}+a_{3n}=\frac1{2n-1}-\frac1{4n-2}-\frac1{4n}=\frac1{4n(2n-1)}$ and we have the partial sums $$S_{3n}=\sum_{k=1}^{3n} a_k =\sum_{k=1}^n\frac1{4n(2n-1)}$$ $$S_{3n-1}=\sum_{k=1}^{3n-1} a_k =\sum_{k=1}^n\frac1{4n(2n-1)}+\frac1{4n}$$ $$S_{3n-2}=\sum_{k=1}^{3n-2} a_k =\sum_{k=1}^n\frac1{4n(2n-1)}+\frac1{4n}+\frac1{4n-2}$$

Stupid question about $1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$

1 Answers1