Integral test | Series | AP Calculus BC | Khan Academy

[@andrewryabchenko2407]

This is great proof of the theorem! The idea of an underestimate of an area is brilliant! Thank you Sal!

[@rishideychowdhury5856]

Thanx. It was really Great Visual Explanation!

[@rintumondal131]

Thanks Khan

[@justinjames8679]

The integral should be from 2 to infinity right? since you included the value for n=1 already

[@sashaburts9676]

Justin James can someone pls answer this??

[@YummyhenIsBeast]

nah, look at the graph and what you're finding is an overestimate. I know you're probably confused by the n=2, but the n=1 accounts for the square with area 1 units^2 while the n=2 and so forth account for the rest of the rectangular areas (1/4, 1/9, 1/16) Essentially, if you were to do 2 to infinity for the integral, it would no longer be a proper overestimate. The overestimate is used to create an upperbound and let you realise that it does not diverge (go to positive infinity) as the series has to be strictly less than 2.

[@subarnasubedi7938]

I was first confused over same thing but its actually correct try cancelling both one's you may realize something!

[@quickmathematics8960]

nice illustration. I liked it

[@mtndewv]

Him: I'm gonna use many colors to make it easier to follow. Me: I'm color blind.. ಠ_ಠ

[@Anonymous-wn6cu]

I am sorry for that

[@Anonymous-wn6cu]

Don’t u have a glasses for this purpose?

[@mtndewv]

@@Anonymous-wn6cu Nah, I wish too expensive and honestly not sure they would work for me b/c of the type of color deficiency (should prob clarify I'm not 100% colorblind just color deficient but the colors he chooses are often hard for me to follow b/c of it. ) ¯\_(ツ)_/¯

[@Anonymous-wn6cu]

@@mtndewv Well, to be honest I don’t have that much of info about this matter. But in case you needed help w/ this video I would be happy to! I will use colors that suits your needs if you wish.

[@mtndewv]

@@Anonymous-wn6cu That is very kind of you but I think all videos would need to be redone in that case lol. The videos are still very helpful I just have a little harder time when it comes to following the colors :) I just thought it was funny when sal said that so I couldn't help but respond. ^-^

[@GogglesPIU]

Why did you use the right hand riemann sum here, but for the diverging series, you used the left riemann sum?

[@smithcodes1243]

SO GOOD.

[@keta7336]

marvelloussssss

[@ozzyfromspace]

You're awesome bro! Just thinking: if the integral is an overestimate, is it possible for a series to be bounded even if the integral on the appropriate domain isn't? Because there is a difference between the curve and the discrete distribution. Isn't this approach assuming that the difference in the domain is finite? I'm not a math person, maybe I'm just thinking too hard. An answer would greatly be appreciated!

[@jameskennedy674]

If you are talking about the curve, I do not think the integral is an overestimate. In this case it is an overestimate in relation to the series being represented. Therefore if the integral + a1 converges then the corresponding series must also converge. The integral without adding in a1 will be an underestimate and therefore if it diverges then the series will also diverge.

[@enigmaticcheers]

great job

[@charleswhitlock4239]

amaze-balls

[@UrDadisMe]

Could you do a video on analysis of the Riemann Zeta function? The sum here is after all a particular zeta value.

[@cboniefbr]

please do the Basel Problem

[@fernandavaxcon]

😢

[@IamCCash]

This is a side of KZbin? Bro.....