Convergence of a series.. Watch

lilman91
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#1
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a_n,b_n>0 with \dfrac{a_{n+1}}{a_n} \leq \dfrac{b_{n+1}}{b_n} for all n\geq 1. If \sum b_n converges, then show that \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..
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Oh I Really Don't Care
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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...
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around
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a_{n+1} \leq a_n \dfrac{b_{n+1}}{b_n}.

Can you see where to go now?
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lilman91
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(Original post by around)
a_{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...
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around
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When faced with a sum it's always a good idea to write out a few terms:

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 a_{n+1} \leq a_n\dfrac{b_{n+1}}{b_n}, we can rewrite the first two terms:

a_1 + a_1\dfrac{b_{2}}{b_1} + a_3 + ...

We can also apply a_{n+1} \leq a_n\dfrac{b_{n+1}}{b_n} inductively: a_{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.
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