If Ʃa_n converges, with a_n > 0, then Ʃ(a_n)^2 always converges

  • Thread starter DotKite
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  • #1
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Homework Statement


if Ʃa_n converges, with a_n > 0, then Ʃ(a_n)^2 always converges


Homework Equations



n/a

The Attempt at a Solution



I am at a complete loss. I have tried using partial sums, cauchy criterion, and I tried using ratio test which seems to work but I am not sure.

Since Ʃa_n converges then by ratio test

lim n->∞ a_n+1 / a_n < 1

Now we apply ratio test to Ʃ(a_n)^2

lim n→∞ (a_n+1)^2 / (a_n)^2 = (lim n→∞ a_n+1 / a_n)^2 < 1^2 = 1

Thus by ratio test Ʃ(a_n)^2 converges.

This working did not utilize the condition a_n > 0, so it seems suspect to me.
 

Answers and Replies

  • #2
jbunniii
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Hint: if ##0 < a_n < 1##, then ##a_n^2 < a_n##.
 
  • #3
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Hint: if ##0 < a_n < 1##, then ##a_n^2 < a_n##.
Ok so you use comparison test right? since (a_n)^2 < a_n and a_n converges then (a_n)^2 converges as well, but is it correct to assume 0<a_n<1? Can you assume that because since
Ʃa_n converges that implies a_n → 0 and for some n≥N a_n is in (0,1)?
 
  • #4
jbunniii
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Ok so you use comparison test right? since (a_n)^2 < a_n and a_n converges then (a_n)^2 converges as well, but is it correct to assume 0<a_n<1? Can you assume that because since
Ʃa_n converges that implies a_n → 0 and for some n≥N a_n is in (0,1)?
Yes, that's right, or putting it more precisely, there exists an ##N## such that ##a_n \in (0,1)## for all ##n \geq N##. And then use the comparison test.
 
  • #5
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thanks a bunch J
 

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