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

[@Jay-no2ql]

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."

[@awowoosas]

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

[@blackpenredpen]

Yup yup!!

[@aashsyed1277]

EASY

[@othman1537]

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

[@manethdulshan9449]

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

[@tgx3529]

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 decreasing.....we can také integral from do And suma=suma1+suma2 ,where suma2 go from no till Infinity.....

[@roshanpoduval8579]

So as to not divide by 0.

[@metegurun5641]

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

[@julietapelletier8570]

maybe I'm late, but the only situation where the integral is ln is when the exponent is -1 (meaning: x^-1 = 1/x ). In this case, the exponent is -1/2, therefore we use the exponential integral, where x^n = (x^(n+1))/(n+1)

[@akunidoreen6503]

What if it is finite case. Does it divergent?

[@abdieloby]

Hi Akuni, I think this is coming 6 months late haha. All finite series converge; a finite sum of finite values will never get to infinity.

[@aashsyed1277]

NO

[@CarlosAlberto-kr8nw]

1/n*sqrt(n) ?????

[@dayanala]

Thank you, Greetings from Honduras.

[@theopapa8232]

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

[@AznJsn82091]

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

[@Vidrinskas]

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

[@rafaeldubois8192]

Tony G Yes, that's the reason why he changed the limits of integration. If he didn't change them, and you plugged the ln(x) instead of the u, you would get the same result.

[@desC0D3]

i used the limit comparison test...

[@lostboy579]

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

[@iamhrz]

Because n>1.

[@helloitsme7553]

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

[@ASYadav-kk2qb]

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

[@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.

[@blackpenredpen]

Konrad Kamiński I would have to first prove that to my students first before I can use it tho. Integral test is fair, not overkill.

[@danielcohen227]

What are your videos supposed to be for in general? Are these actual college classes or tutoring help?

[@blackpenredpen]

Daniel Cohen mainly for my students who I teach at a college. But of course, it's for everyone who finds them helpful

[@danielcohen227]

Very nice lessons! I follow you very well!

[@MShazarul]

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

[@blackpenredpen]

Yup!!

[@wellpickup4186]

If n=1 then the given series div

[@siddartharcot362]

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

[@hefzibac286]

👌

[@gmedran3]

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

[@blackpenredpen]

Yup

[@boredgamesph4872]

0! = 1

[@julietapelletier8570]

@@boredgamesph4872 LOL

[@meamarie8714]

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

[@Frank-xc8ys]

Si tan solo hablaras español .....lamentablemente es asi

[@yaserrivera5565]

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

[@CmG626xX]

what vichet said hahaha

[@blackpenredpen]

Lol!! Yes

[@rafaelv.t1403]

if n=1 ln (1) is undetermined Because E^Anything is not 1 (I Am 6 Grade )

[@Fokalopoka]

e^0 is 1, its that you get division by zero if n=1