Find the series $\sum_{n=1}^{\infty}{a}_{n}$ if ${s}_{n}\in \mathbb{N}$ is ${s}_{2n-1}=1$ and ${s}_{2n}=0$

Question

Find the series $\sum_{n=1}^{\infty}{a}_{n}$ if ${s}_{n}\in \mathbb{N}$ is ${s}_{2n-1}=1$ and ${s}_{2n}=0$

I used ${a}_{n}={s}_{n}-{s}_{n-1}\Rightarrow {a}_{2n}={s}_{2n}-{s}_{2n-1}\Leftrightarrow {a}_{2n}=-1$ but it doesn't seem to end up anywhere. There must be a way to find ${a}_{n}$ so I can find the series(I mean write the formula). Any suggestions?

Thank you.

edit: The answer is $\sum_{n=1}^{\infty}{(-1)}^{n-1}$ but i don't know how we get there.

Answer 1

If $s_n$ denotes the the sequence of partial sum then the series will not be covergent. As the series will be covergent if and only if the sequence of it's partial sum converges. Here two subsequence of the sequence of partial sums converge to two different limits. So clearly the sequence of partial sum will not converge.

Answer 2

You're on the right track:

$a_1 = s_1 = 1$

$a_2 = s_2-s_1 = -1$

$a_3 = s_3-s_2 = 1$

$\dots$

$a_{2n-1} = s_{2n-1} -s_{2n-2} = 1$

$a_{2n} = s_{2n} -s_{2n-1} = -1$