integral test, series of 1/(n*sqrt(ln(n))), calculus 2 tutorial

Does this Series Converge or Diverge? Series of 1/(n*sqrt(ln(n)))

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#blackpenredpen #math #calculus #apcalculus

[awowoosas]

When n = 1, you will have ln(1) = 0 in the denominator.

[Jay-noql]

My course had me have to use the limit comparison test! I kept getting 0, which is inconclusive, for every series I chose. It said literally "you may only use the LCT."

[manethdulshan]

Nice
I got the idea to solve 1/[n.(Ln(n))^p]
Same as this method
Thank you 🥰

[othman]

you can also see (1/x) * ln(x)^(-1/2) as the derivative of 2 * ln(x)^(1/2) directly, without using the change of variable

[tgx]

Yes, Great!
I think, we can use also information, there lim f(x)=0, where x go till Infinity, f( x) Is not negative, So exists xo, the function f(x) will be for x>xo can také integral from do And suma=suma1+suma2, where suma2 go from no till

[lostboy]

I really needed to see this literally the question i couldn't finish on my test 🙏🏽 way to clutch up

[AznJsn]

In the improper integral, once we take everything to the u-world, we could just stop and do the p-series test, to say that p=1/2<1. So that improper integral diverges, without needing to evaluate all the way through

[theopapa]

because the first term for n = 1
jt would be 1/0

[siddartharcot]

if u do the integral test with sigma from 1 to n=infinity for n which is just 1+2+3+4+... it will say the definite integral is infinity, therefore does diverge however we know it actually converges

[helloitsme]

I could sense the diverge cause 1/1+1/2+1/3.. Diverges and √(lnx) is so small it barely makes a difference. Still doesn't proof it tho but cool to feel a diverge/converge coming

[LeBartoshe]

It would be way simpler and faster to use Cauchy condensation test.







Just saying xD

[kokainum]

I think you overkill it. Simpler thing would be to use criterion for decreasing positive sequences that says that sum a_n converges if and only if sum (2^n *a_(2^n)) converges. It's useful for these series with logarithms.

[ASYadav-kkqb]

Un= (n-1)/√log n what method i should apply to check series is convergent or divergent??

[akunidoreen]

What if it is finite case. Does it divergent?

[MShazarul]

Because ln(1) = 0, so as to not have the denominator as 0.

[Vidrinskas]

Hmmm, but if the integral diverges for u does that mean it automatically diverges for x?

[gmedran]

if N started at 1, the denominator is 0! So the Integral test would give you no help!

[metegurun]

Hey man at 4:44 why didn't u just integrate 1/ u^1/2 as ln (u^1/2) ?

[meamarie]

wow I haven't seen a chalkboard in a long time

[yaserrivera]

O mi inglés está mejorando o las matemáticas te permiten entender que coño hace otra persona en otro idioma