My Math Forum Sequences

 Real Analysis Real Analysis Math Forum

 May 30th, 2017, 07:46 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Sequences Consider the real sequence $(a_n)$ and $(b_n)$ such that $\displaystyle \sum a_n b_n$ converges. Which of the following statements is true ? A) If $\displaystyle \sum a_n$ converges, then $(b_n)$ is bounded. B) If $\displaystyle \sum b_n$ converges, then $(a_n)$ is bounded. C) If $(a_n)$ is bounded, then $(b_n)$ converges. D) If $(a_n)$ is unbounded, then $(b_n)$ bounded. I'm guessing it as option A. Please check and let me know Thank you
May 31st, 2017, 05:07 AM   #2
Senior Member

Joined: Nov 2015

Posts: 232
Thanks: 2

Quote:
 Originally Posted by Lalitha183 Consider the real sequence $(a_n)$ and $(b_n)$ such that $\displaystyle \sum a_n b_n$ converges. Which of the following statements is true ? A) If $\displaystyle \sum a_n$ converges, then $(b_n)$ is bounded. B) If $\displaystyle \sum b_n$ converges, then $(a_n)$ is bounded. C) If $(a_n)$ is bounded, then $(b_n)$ converges. D) If $(a_n)$ is unbounded, then $(b_n)$ bounded. I'm guessing it as option A. Please check and let me know Thank you

 May 31st, 2017, 05:24 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,598 Thanks: 2583 Math Focus: Mainly analysis and algebra If A is true, then B must also be true. I think, from memory, that both are. If $(a_n)$ is bounded, but not convergent, then $(b_n)$ must converge to zero. This is a necessary condition but not sufficient. Similarly, if $(a_n)$ is unbounded, then $(b_n)$ must converge to zero. This is a necessary condition but not sufficient. Last edited by v8archie; May 31st, 2017 at 05:29 AM.
 June 1st, 2017, 05:20 PM #4 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 I have one more doubt. Can you please let me know the difference between Option A and Option C ?
 June 1st, 2017, 05:29 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,598 Thanks: 2583 Math Focus: Mainly analysis and algebra In option C, the sum $\sum a_n$ doesn't necessarily converge. In fact, the sequence $(a_n)$ doesn't even have to converge $a_n = \sin n$ would be one such sequence. Equally, although the sequence $b_n$ converges, the series $\sum b_n$ doesn't necessarily converge even if $(b_n)$ converges to zero. e.g. $b_n = \frac1n$. Looking again at the question, given that $\sum a_n b_n$ converges then A and B can be false. For example: $a_n = \frac1{n^3},\, b_n = n$. C can also be false using the same example. D is true, because if the sequence $(a_n)$ is not bounded, the series $\sum b_n$ must converge quickly enough to make the sum $\sum a_n b_n$ converge. An example of this is $a_n=n,\, b_n=\frac1{n^3}$. Last edited by v8archie; June 1st, 2017 at 05:40 PM.

 Tags sequences

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post DATAfiend Pre-Calculus 3 November 12th, 2014 03:17 PM alexpasty Calculus 2 April 24th, 2013 08:20 AM ely_en Algebra 1 November 11th, 2011 02:07 PM Kiranpreet Algebra 1 November 11th, 2008 03:48 PM cindyyo Algebra 2 August 20th, 2008 03:40 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top