Calculus II - Series and Convergence

GreenPrint
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Homework Statement



Determine if the series

inf
Sigma n/(2n+1)
n=1

converges

Homework Equations





The Attempt at a Solution



When i did this I originally I thought I would just apply the divergence test

lim n/(2n+1) =/= 0
n->inf

there fore I thought by the divergence test the series diverges but I guess sense the limit is defined to be 1/2 it converges...

I'm confused... I think I may be over complicating this but by the divergence test shouldn't this series diverge sense the limit does not equal zero?

Thanks for any help
 
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GreenPrint said:
... but by the divergence test shouldn't this series diverge sense the limit does not equal zero?

Yes, that's correct. It's more commonly called the nth term test for divergence.
 
GreenPrint said:

Homework Statement



Determine if the series

inf
Sigma n/(2n+1)
n=1

converges

Homework Equations





The Attempt at a Solution



When i did this I originally I thought I would just apply the divergence test

lim n/(2n+1) =/= 0
n->inf

there fore I thought by the divergence test the series diverges but I guess sense the limit is defined to be 1/2 it converges...

I'm confused... I think I may be over complicating this but by the divergence test shouldn't this series diverge sense the limit does not equal zero?
Thanks for any help

The bolded part is correct. Since you took the limit of the summand and received 1/2, the limit diverges. If the sum converges its limit is equal to 0.

However, the converse of the statement is not true. Don't fall into the trap of thinking the sum converges if the limit is equal to 0. If you obtain a limit of 0, you need to resort to a different method to determine if the sum converges.
 
hm thanks i think i got it mixed up and it asked me if the sequence converges which it does lol
 
GreenPrint said:
hm thanks i think i got it mixed up and it asked me if the sequence converges which it does lol

No it doesn't. It diverges. You just showed this using your test. Or was that a typo?

And it's a series, not a sequence.
 
ya i got it and understand now i got a different part mixed up and stuff
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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