The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Convergence of a series..

an,bn>0a_n,b_n>0 with an+1an≤bn+1bn\dfrac{a_{n+1}}{a_n} \leq \dfrac{b_{n+1}}{b_n} for alln≥1 n\geq 1. If ∑bn\sum b_n converges, then show that ∑an\sum a_n converges.

I was thinking of using ratio test conditions here, btu the that sort of falls apart if we have the b_n series converging but the ratio tending to 1. Any ideas how I should go about this..
There is the other fact that a series sum a_n can converge while the limit a_n+1 / a_n dosen't exist.

If sum b_n converges then b_n tends to 0...
Reply 2
Avatar for around
around
13
an+1≤anbn+1bna_{n+1} \leq a_n \dfrac{b_{n+1}}{b_n}.

Can you see where to go now?
Reply 3
Avatar for lilman91
lilman91
OP
4
around
an+1≤anbn+1bna_{n+1} \leq a_n \dfrac{b_{n+1}}{b_n}.

Can you see where to go now?


Unfourtunately, not really. I know i need to get an upper bound on a_n to do with b_n or something about b so that someting to do with the b series bounds the a series and so it converges.

I'm not sure because as Deank22 said, b_n+1/b_n may not converge and so it may not be bounded...
Reply 4
Avatar for around
around
13
When faced with a sum it's always a good idea to write out a few terms:

a1+a2+a3+...a_1 + a_2 + a_3 + ...

We know b_n converges, so it's a good idea to somehow bound the terms of our sequence above by the terms of the b_n sequence. Using the fact that an+1≤anbn+1bna_{n+1} \leq a_n\dfrac{b_{n+1}}{b_n}, we can rewrite the first two terms:

a1+a1b2b1+a3+...a_1 + a_1\dfrac{b_{2}}{b_1} + a_3 + ...

We can also apply an+1≤anbn+1bna_{n+1} \leq a_n\dfrac{b_{n+1}}{b_n} inductively: an+2≤an+1bn+2bn+1≤anbn+2bna_{n+2} \leq a_{n+1}\dfrac{b_{n+2}}{b_{n+1}} \leq a_n\dfrac{b_{n+2}}{b_n}.

You can finish this off now.

Related discussions

Latest

Trending

Trending

Articles for you