Finding a limit using an integral!

[@nachiketagrawal5154]

I've found your selection of problems very interesting. Also very in depth, but not boring style of teaching is very effective. Hats off. You might wanna try the Indian National Mathematics Olympiad questions. They're pretty solid and cover a vast variety of topics like algebra, geometry, inequalities, number theory. Great to see you getting recognition.

[@LucasEccard]

You Could also use Stolzs Cèsaro's theorem ( L'Hopital for somatories) where lim f(a)/g(b) = f(a+1)-f(a)/g(a+1)-g(a) (In the cases of infinity/infinity only). Just doing that 2 times and opening the first terms of the binomial gives the answer : )

[@reginalmuonaka2297]

Man, you give me hope again. May God bless you.

[@goodplacetostop2973]

6:39 almost

[@sakshamgarg8531]

a= -1,-2 : they WON'T work.(because then the intgration would result in ln(x) in denominator or the numerator.). btw nice video. happy to have to my concepts revised and also thrilled to solve them. keep uploading ,i have subscribed your channel ;).

[@pengiiin]

Though the limits still exist: a=-1: we get (1+...+1)/(n*H_n)=1/H_n, where H_n is the harmonic series up to 1/n. This series diverges, therefore the limit is 0. a=-2: we get H_n/(n*(1/1²+1/2²+...+1/n²)). The sum of the reciprocals of the squares converges, while H_n/n goes to zero fast (H_(2^n)

[@Czeckie]

the question about correct a's seems rather subtle: the calculation doesn't work for any a

[@williamsean3469]

I guess it should be pointed out that when a= -2 the original value becomes the sum of 1/n^2 in the denominator which is convergent and the top becomes the harmonic series which is approximately ln+1) but becomes zero once divided by n. When a is less than -2, say -3 both the numerator and the denominator converges, but becomes zero once divided by n.

[@sandorszabo2470]

As an alternative calculation is, without Riemann sums, using the en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem .

[@danielsin5563]

Subscribed!

[@jogeshgupta7583]

He is such a brilliant mathematician.....looks as if he solves everything by formulas which are a part of his intuition ......His subscribers should increase at an exponential rate y=A(1-e^-kx).... but the graph should always go up

[@m4riel]

I kinda cheated using the pattern that the sum of the integers from 1 to n raised to the a-th power has a leading coefficient of n^(a+1)/(a+1), then it just ended up being n^(a+2)/(a+2) * (a+1)/n*n^(a+1)

[@rrr1304]

In iit this type is favourite.

[@nontth5355]

a = -1 wont work too I think Because int (1/a) da = ln(a) + C

[@nontth5355]

@Joao Vitor bruh I feel so dumb now xD How can I forgot that case...

[@nournote]

It would be more interesting to cite used theorems, unless you're a physicist. Subscribed.

[@MichaelPennMath]

I am kind-of a mathematical physicist: aip.scitation.org/doi/full/10.1063/1.5045164

[@rrr1304]

Sir please try jee advanced 2016 question something like that but little hard . May be in paper 1 or 2.

[@Mathelite-ii4hd]

perfect!:)