On the conjecture that $1-\frac 13 + \frac 16 +\frac 1{10} -\frac 1{15}+\cdots = 1\frac 19$

Question

Conjecture:

$$1-\frac 13 + \frac 16 +\frac 1{10} -\frac 1{15}+\cdots = 1\frac 19$$ where the pattern of the signs is $+,-,+,+,-,+,+,+,-,\cdots$ and the denominators are the triangular numbers.

Whatever this series converges to (if it does), it converges so slowly. I was on my calculator manually doing this for hours (eventually using two at once) and, unless I have erred somewhere, it seems this approaches $10/9$.

Given the pattern of the signs, I don't think there is a way to write this using summation notation. If I could, then I'd be going straight to Wolfram Alpha. But can this series be shown either convergent or divergent only with by-hand calculation?

Thanks.

Answer 1

Your series can be written as \begin{align} S&=1-\frac 13 + \frac 16 +\frac 1{10} -\frac 1{15}+\cdots \\ &=\left( 1+\frac 13 + \frac 16 +\frac 1{10} +\frac 1{15}+\cdots\right)-2\left(\frac 13+\frac 1{15}+\cdots\right) \end{align} where both series in the expression converge, first one is equal $2$ (sum of reciprocal triangular numbers), second one is $\frac{4}{9}$ (each term is reciprocal of $n(n+3)/2$-th triangular number). Algebraically $$ S_1=\sum_{n=1}^{\infty}\frac{2}{n(n+1)}=2,S_2=\sum_{n=1}^{\infty}\frac{8}{n(n+1)(n+2)(n+3)}=\frac{4}{9}, $$ both series can be computed by telescoping (see also Find sum of infinite anharmonic(?) series). So the result is $$ S_1-2S_2=2-2\frac{4}{9}=\frac{10}{9}. $$

Answer 2

The absolute values of the terms are dominated by 2 / n ^ 2 (2 > 1, 1/2 > 1/3, 2/9 > 1/6, and so forth). The summation of 2 / n ^ 2 is pi ^ 2 / 3 (see https://en.wikipedia.org/wiki/Basel_problem). So, it is convergent.