Does converging imply is bounded?
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No, take . It's true if though.
Originally Posted by Bruno J. No, take . It's true if though. Whoops, I meant to say . Why is this true?
Last edited by chiph588@; May 19th 2010 at 07:30 PM.
Originally Posted by chiph588@ Whoops, I meant to say . Why is this true? I'm sorry, it's false! Take ... It's true if and converges monotonically.
Originally Posted by Bruno J. I'm sorry, it's false! Take ... It's true if and converges monotonically. Take 1/(n+1) < 1/n => 2 - 1/(n+1) > 2 - 1/n Here , is strictly monotone converging to 2. So, and not 0. For a convergent series we need that the tail end of the series go to 0.
So it should be false , but can we find all the sequences that are the counterexamples for this statement ?
I just gave you one such example. For any C > 0 , take converges to C , however is not convergent.
Originally Posted by ques I just gave you one such example. For any C > 0 , take converges to C , however is not convergent. I was asking whether the sum is bounded, not convergent.
Originally Posted by ques I just gave you one such example. For any C > 0 , take converges to C , however is not convergent. What chip said! The partial sums you give are bounded. To prove it, let where . By the alternating series test, converges, which immediately implies the partial sums are bounded.
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