2
$\begingroup$

Consider the series $\displaystyle\sum^{\infty}_{n = 1} a_n$ where $a_n \geq 0$ and the related series $\displaystyle\sum^{\infty}_{n = 1} A_n$ where the $A_1 = \displaystyle\sum^{n_1}_{n=1}a_n, A_2 = \displaystyle\sum^{n_2}_{n=n_1+1}a_n, \ldots, A_n=\displaystyle\sum^{n_N}_{n=n_N+1}a_n$. Prove the series $\displaystyle\sum^{\infty}_{n = 1} a_n$ converges iff $\displaystyle\sum^{\infty}_{n = 1} A_n$ converges.


I was thinking that if I could show $\displaystyle\sum^{\infty}_{n = 1} a_n = \displaystyle\sum^{\infty}_{n = 1} A_n$ then it would show they converge as I would assume that $\displaystyle\sum^{\infty}_{n = 1} a_n$ converges but I am not sure how to start it. Could anyone point me in the right direction?

$\endgroup$
  • $\begingroup$ What do you know about the sequences of partial sums of both series? $\endgroup$ – Daniel Fischer Oct 4 '16 at 19:14
  • 3
    $\begingroup$ The second series is just grouping the terms differently. $$\sum_{i=1}^{n_k} a_i = \sum_{i=1}^k A_k$$ $\endgroup$ – Alexis Olson Oct 4 '16 at 19:18
  • $\begingroup$ So it is not completely correct, try with $a_n = (-1)^n$, $A_n = a_{2n}+a_{2n+1} = 0$ $\endgroup$ – reuns Oct 4 '16 at 19:19
  • $\begingroup$ Certainly not "iff". Let $a_n=(-1)^n$ and $n_k=2k$. $\endgroup$ – Hagen von Eitzen Oct 4 '16 at 19:20
  • 1
    $\begingroup$ @user1952009 the OP stated that $a_n \geq 0$. $\endgroup$ – Mosquite Oct 4 '16 at 19:21

Your Answer

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

Browse other questions tagged or ask your own question.