# prove or disprove that $\sum (a_{2n-1} + a_{2n}) = a_1 + \sum (a_{2n} + a_{2n+1}) = S \Longrightarrow \sum a_n = S$

$$(a_n)_{n=1}^\infty$$ is a sequence, and it is known that that $$\sum_{n=1}^\infty b_n = (a_1 + a_2) + (a_3 + a_4) + ... = S$$ and $$\sum_{n=1}^\infty c_n = a_1 + (a_2 + a_3) + (a_4 + a_5 + ... = S$$ I need to prove (or disprove) that $$\sum_{n=1}^\infty a_n$$ converge and $$\sum_{n=1}^\infty a_n = S$$. Because $$\sum_{n=1}^\infty b_n$$ and $$\sum_{n=1}^\infty c_n$$ converge, $$\sum_{n=1}^\infty 2\cdot c_n - b_n$$ also converge and I tried to do this $$\sum_{n=1}^\infty 2\cdot c_n - b_n = 2a_1 - (a_1 + a_2) + 2(a_2+a_3) - (a_3+a_4)+ ... = a_1 +a_2 +...=\sum_{n=1}^\infty a_n=S$$ and so $$\sum_{n=1}^\infty a_n$$ converge, but then I thought that if $$a_n = (-1)^n$$ then both $$\sum_{n=1}^\infty b_n$$ and $$\sum_{n=1}^\infty c_n$$ converge but $$\sum_{n=1}^\infty a_n$$ obviously doesn't, hence what I wrote before is wrong (but doesn't disprove it because $$\sum_{n=1}^\infty b_n = 0 \neq -1 = \sum_{n=1}^\infty c_n$$), and now I have no idea how I can prove it, and all the examples I can think of support the statement.

EDIT: I tried to use Cauchy convergence test to test if $$\sum a_n$$ converge, but unless $$a_n \to 0$$ (which isn't given) it doesn't seem to work

• Hint: According to definition, we need to prove $\forall\epsilon,\exists N,\forall n>N,\sum_n a_k<\epsilon$. Then consider $N_1,N_2$ of the two series. – Yao Fu May 2 at 16:35

Every partial sum $$\sum_{n=1}^M a_n$$ of the $$a$$ series is also a partial sum of the $$b$$ series or the $$c$$ series (depending on whether $$M$$ is even or odd) and will therefore be close to $$S$$ once $$M$$ is large enough.
• but this is disproven by $a_n=(-1)^n$ each partial sum of $a_n$ is either $(-1)$ or $0$ but all partial sums of $b_n$ are $0$ – CforLinux May 2 at 17:01
• @CforLinux If $a_n=(-1)^n$ then the first equation in the hypothesis says $S=0+0+0+\dots=0$ and the second says $S=(-1)+0+0+\dots=-1$, so the hypotheses of the problem are not satisfied. – Andreas Blass May 2 at 17:08