I was struggling with a similar problem, this video helped me a lot. Thank you.

LCT is just a more formal version of the Squeeze Theorem, no? I was just thinking along the lines of: For all x in (0,1], sin x < x. since each term of 1/n^2 is an element of (0,1], then sin(1/n^2)

“Which do you like better?” Both are very useful for different situations. So I don’t have a preference.

I have come up with a similar test for divergence for definite integrals with a singularity at some value. You take the limit as some variable, a, goes to 0 of the area between n and n+a where f(n+a) diverges. That area is like a thin, tall rectangle so its the limit of a*f(n+a). If its zero, it may or may not converge, otherwise it diverges

It is also well known that |sin(x)|

My test about series is today :p a video like this is refreshing

Couldn't have made it any easier, thank you professor!

At 5:40 you could just use limit (sin(a)/a), while a->0 instead of using Lopital's rule :-)

Your videos are amazing. Congrats from spain :)

Have you tried using double sided markers to use up to four colors?

Best math videos on youtube for calc bc no questions 👍🏽

For the first sum one one could say that on the open interval (0;pi/2) 0

where can I find the microphone you are using? May you do some video about the Lebesgue-integral or measure theorie as well?

cos(x) = 1 - x^2 ... immediately obviously divergent. (cos(1/n^2)-1) definitely converges tho, same with sin(1/n^2). Because (cos(x)-1) and sin(x) has form x^k only, (1/n^2)^k only makes each term smaller.

pls do a video on what it converges to

What if I had infinity product for cos 1/n²?

But what is the actual sum if it does converge?

Amazing!

Can't you just say that as n approaches infinity lim(sin(1/n^2)) ~ lim(1/n^2) or does that only work if both the numerator and the denominator are infinitesimals?

Your video is Good, I like your video!!

Muy bueno! greetings from Uruguay

Chen lu!

hey professor, i recently calculated the derivative of arctanh(sin(x)) and i realized it was sec(x). Does this mean that arctanh(sinx) is the same as ln(secx+tanx)? I plugged it into a graphing calculator and they were the same, except that arctanh(sinx) has extra "lines". Why does this happen?

Doens't both series have the same value? I mean, aren't they the same thing but backward? I tried to integrate and it seems like they have the same value

i think it would be that little bit easier if we just let 1/n^2 to be u , as n goes to infinity , u goes to 0 and by the limit sin(0)/0 =1

8:29 writes something while saying something else ... That's clever, I've never been able to do that :)

How about using Taylor expansion?it's the same

One year ago you solved the integral of sqrt(tan(x)). Could you solve the integral of tan(sqrt(x))????

Why cannot we draw any conclusion if lim n-> infinity sin(1/n^2) is 0?

Now that you’ve shown Σsin(n^-2) is convergent, care to do a video on evaluating it?

Cool. What I thought was that for x≥1, sin(1/x²) is strictly decreasing and also bounded as such: 0

i was watching a video and i got curious, there is any irrationals numbers that sum equals an rational number ?

Why does changing it to the x-world matter? isn't that just a personal preference? You can treat "n" as the independent variable and do d/dn instead of d/dx. Right?

Could also have presented sum from n=1 to infinity of tan(1/n^2) since it's on the same kind of exercises. It also converges.

I watch this video before I go to school.

Im gonna keep it a hunnid i fw the LCT more than the LFD. chen lu!

How do I take the limit of sin(1/x) as x approaches 0?

The LCT test is for positive series, how do you know that sin(1/n^2) is positive to every natural number?

1/(π^n)-(n^π) converge????

Pardon me Professor, may i ask a question ? Which is : One the left side equation, when you subtitute 1/(n^2) = x , x should be approaching to 0 while n^2 is going to infinity. I think you should write that way. That's all professor. But your whole answer completely right. Correct me if i am wrong. Hope you succes Professor ☺☺☺👍👍👍Thank You

But if cos(x^-2) is < x^-2 for all x greater than ≈ 0.739085 and x^-2 converges from 1 to inf shouldnt cos(x^-2) do the same? Edit: nvm

Use l Hospital rule for calculate the limite that defined sin'(0).... Strange

I prefer BOTH!!!

Hello sir your voice is very cute and teach very good

those happy and sad faces ❤

The first is between 1 and pi^2/6 and the second diverges :p

did you just say Chen Lu

1≠ ☺

5:27 Just do x=1/n^2 Then you get: lim(x->0) sinx/x = cosx = cos(1/n^2)

Not Chen Lu😂😂😂

sin(x)

8:40 midlife crises

👍👍👍👍

I don´t even understand you tshirt :,v

1:25 well duh

Leibniz is the best one

(x-1)^x=x^(x-1) Can anybody solve that?

5:20 Can't we evaluate this limit more simply by letting t = 1/n^2. Now t goes to zero as n goes to infinity, so equivalently we have the limit as t goes to zero of sin(t) / t = 1

Why is 7 afraid of 8? Because 7, 8, 9. Guy, I'm suppoused to be doing my homework. I saw this eating a mango. Have you eat it?

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