#convergence or divergence

27 messages · Page 1 of 1 (latest)

obtuse flax
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Test the series whose general term is :

(n²+1)^½ - n

graceful tundraBOT
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obtuse flax
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So according to me it's convergent but the solution refers this as a divergent series

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Take n² common and out of the parenthesis and then expand the binomial expression. The result can be simplified to form the eq:

1/2n -1/8n³+1/16n⁵....
This should be a converging series right ?

stark canopy
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Just to verify, is this what the general term is?

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Because I am a bit confused

craggy current
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@obtuse flax use the integral test to verify that it diverges

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Verify that it is a monotonically decreasing function. Idk if its easy to find the integral but it works.

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Maybe some other convergence test could be used as well.

frosty dirge
obtuse flax
frosty dirge
craggy current
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@obtuse flax ok what about the limit comparison test where an=√(n²+1)-n and bn=1/n. You know that both an, bn are positive and you have the limit n→∞ an/bn as n→∞ n(√(n²+1)-n)

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Multipling and dividing by √(n²+1)+n the limit becomes n/(√(n²+1)+n) pull out n from denominator and numerator you have 1/(√(1+1/n²)+1) which for n→∞ has limit 1/2. Therefore the limit exists and it is finite non zero. Since the series of bn=1/n (harmonic series) diverges then the series of an=√(n²+1)-n diverges as well

frosty dirge
obtuse flax
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Ah yes I know the comparison test using an auxiliary series

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I'm sorry I found this confusing too but I understand a lil what you're explaining

craggy current
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Well if you write it down on paper, i guess it will be more understandable

obtuse flax
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Ohh okayy I understand now!

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Thanks a lot @craggy current and @frosty dirge !!!

rough forumBOT
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@obtuse flax has given 1 rep to @frosty dirge @craggy current

frosty dirge
obtuse flax
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Still I appreciate your time and attention