2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.