$(a_n)_{n=1}^\infty$ 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

  • $\begingroup$ 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. $\endgroup$ – 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.

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  • $\begingroup$ 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 $0$ $\endgroup$ – CforLinux May 2 at 17:01
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    $\begingroup$ @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. $\endgroup$ – Andreas Blass May 2 at 17:08

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