This infinite series is crazy!

We look at a nice infinite sum.

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[hanibahout]

Nice Infinite sum. Thank you Michael. Merry Christmas.

[amirb]

5:53 note 2^{-x} \neq 2^{1/x} but the original expression on the left of blackboard was correct

[nathanisbored]

**writes (1 - 2ⁿ√2 without closing it multiple times**
"ya that seems like a good place to stop"

[Stelios]

Even Putnam-alike or IMC-alike series may be evaluated through telescoping summation! Here we saw that one more crazy series was killed once telescoping summation came into play. Never underestimate the power of telescoping summation!

[gonderage]

"i think that's a good pla-" and the video immediately stops

[davidgillies]

That's cool. I remember working through the "fact" with my daughter a couple of years ago for her 2nd year calculus class. It's a neat thing to have in your toolbox.

[danielmilyutin]

There was no need for second tool at all. Just replace 2^-N with x->0.

[davidbrown]

Hats off to you Sir...huge respect...you are one heck-of-a clever dude.

[urumomaos]

You can also replace 2^n with 1/x, as x goes to 0. You then are left with x/(1-2^x) which clearly has limit -1/ln2

[manucitomx]

Merry Xmas, professor! Thank you for sharing these.

[abc]

I like the conviction and gravity with which you articulate. You can be good teacher.

[tomatrix]

Great. Happy christmas to you and your family Michael, stay safe!

[johnsalkeld]

NICE - IT CAN BE USED AS PART OF ESTIMATION THEORY - HENCE OR OTHERWISE SHOW THAT EXP(X) CROSSES from 1.xx to 2.xx when x is between 2/3 and 3/4

[carlfriedrickgauss]

I have a problem suggestion,
Find all functions f:N-->N such that n!+f(n)+m divisible by f(n)!+f(m+n) for all m, n in N.(0 is not an Natural Number.)

[Meverynoob]

Yo someone else pointed it out but big oof with the power there, a^(-b) is 1/(a^b) and not a^(1/b) which is the bth root of a. Very big mistake, not sure if it ruins the entire solution

[arimermelstein]

I wonder why when you do partial fractions, you always create a system of equations out of the terms, but there’s an easier way when there are linear terms.
In this case, if we plug in u= -1 you cancel the B term and plug in u=1 and you cancel A.

[arijitdeka]

That was quite informative, Michael. If possible, could you please take some problems involving limits of infinite series?

[parameshwarhazra]

Sir I am looking for these kind of crazy integrals, thank you. With ❤️ from India 🎉🎉🎉

[tomasstride]

Unusually I have had a couple of good days. I was able to do yesterday's problem just by looking at the thumbnail and not even starting the video. Today I also did the question but I did start the video and look at the very opening. The thing is having seen the results to be used it is fairly easy to see what is to be done. My self doubt comes from the thought that had I not seen which results were to be used I probably would not have come up with them myself.

[tutordave]

At about 6:03 I notice a mistake b/c you wrote 2^(-x) as 2^(1/x). But you fixed it in all later steps.