#convergence or divergence
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So according to me it's convergent but the solution refers this as a divergent series
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 ?
@obtuse flax use the integral test to verify that it diverges
Verify that it is a monotonically decreasing function. Idk if its easy to find the integral but it works.
Maybe some other convergence test could be used as well.
im pretty sure u can use the integral test here
oh yup
Yess
Oh I haven't learnt it yet
if u know what integrals r its rlly simple
@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)
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
if u dont use latex i am mentally unable to understand what ur saying but yh ig if u find the right sequence i would work
Ah yes I know the comparison test using an auxiliary series
I'm sorry I found this confusing too but I understand a lil what you're explaining
Well if you write it down on paper, i guess it will be more understandable
@obtuse flax has given 1 rep to @frosty dirge @craggy current
i mean i didnt necessarily help much but ok lol
Still I appreciate your time and attention