0
$\begingroup$

We can see the fact that:

If a series $\sum_{n=1}^{\infty} a_{n}$ converges then:

$\displaystyle \lim_{n \rightarrow \infty} (a_n + a_{n+1} +···+ a_{n+r} )=0 $

This is my proof:

$\displaystyle \lim_{n \rightarrow \infty} (a_n + a_{n+1} +···+ a_{n+r} )$

$=\displaystyle \lim_{n \rightarrow \infty} a_n + \displaystyle \lim_{n \rightarrow \infty}a_{n+1} +···+\displaystyle \lim_{n \rightarrow \infty} a_{n+r} $

$=0+0+...+0=0$

Is it correct?

Also I want to ask:Does the converse of the implication holds:

That it: Does $\displaystyle \lim_{n \rightarrow \infty} (a_n + a_{n+1} +···+ a_{n+r} )=0 $ imply the series $\sum_{n=1}^{\infty} a_{n}$ convergent?

Whether it is true or not. I am searching for a proof and a justification. Could someone help to prove or disprove the statement?

Thanks so much !

$\endgroup$
1
$\begingroup$

As long as $r$ is finite, I believe your answer is correct. The converse is not true. Let's, for example, let $a_n = 1/n.$

$\endgroup$
0
$\begingroup$

The idea you used in your argument is correct, yet you need to express it in a more rigorous way.

The inverse implication is not true. We can particularly consider the sum where we have only one term. We know that the terms of the harmonic series converges to 0 where the sum is divergent.

$\endgroup$
  • $\begingroup$ Em...If it is not rigorous enough. May I ask you to show how to write it precisely and rigorously? Thanks! $\endgroup$ – PropositionX Oct 7 '16 at 21:15
  • 1
    $\begingroup$ Maybe state that the first equality follows because you have a finite number of terms, and each one has a limit. The second one comes from the term test. $\endgroup$ – trang1618 Oct 7 '16 at 21:52

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.