Math for fun#15, GEOMETRIC SERIES WANNABE?

Math for fun#15, GEOMETRIC SERIES WANNABE?

blackpenredpen
math for fun,

tags: math for fun has hard math problem, challenging math problems, hard algebra problems, hard calculus problems, sum of b^ln(n), series b^ln(n), hard series problem, calculus 2 series problems, find the values of b so that the series b^ln(n) converges, blackpenredpen, air jordan 11s space jam on feet, jordans 11s space jam, space jam in the background.

[jameswong]

if b = e, then e^(ln(n)) will become n, and the sum of all positive integers is -1/12 :D #totallylegit #mathgenius

[mbabb]

Really loving your "Math for Fun" series! Thank you for the great videos and Happy Holidays!!!

[martind]

In fact surely we can generalise this to any log base: Sum of b^log_x(n) converges when 0 < b < 1/x?

[qq]

Why can't you do the comparison test to the geometric series? You will find that the solutions are 0<b<1, not just 0<b<1/e

^Note that 1/e is less than 1

[pierrecurie]

The limit b->0 exists and is 0. Furthermore, this thing converges for -1/e<b<0 (but the result depends on choice of branch cut).

This sum simplifies to Zeta(Ln(1/b)).
The convergence condition is then Re(Ln(1/b)) > 1.
This can then be rearranged to abs(b)<1/e

[JACdUpProductions]

that was such an excellent breakdown, you turned something that initially boggled my mind into something i could actually see as making a lot of sense with math manipulation. thank you so much!

[TheLucidDreamer]

1/e is the harmonic series, which is known to diverge

[someperson]

ln(b) < -1 implies b < e^(-1) because e^x is strictly increasing. Of course, your comment "ln(x) is strictly increasing" implies that the inverse of ln(x), (i.e., e^x) is also strictly increasing. Also, "SIGMA(1/n^p) converges iff p > 1" (i.e., "iff" instead of "if") would improve the logic of the argument. Thanks for your excellent videos.

[JianJiaHe]

b^ln(n) = n^ln(b), so 0 < b < 1/e.

[richardaversa]

interesting solution! I'm wondering, couldn't you just do the ratio test (<1 for convergence)?

[leaolp]

This is Magic! I have aa Signal and Systems Exam tomorrow and that just helped me! Thanks again!

[timluyten]

Hey man, love your videos! I was wondering where you teach mathematics

[VaradMahashabde]

(negative)^irrartional can be defined with complex numbers
x is any number
Let n = -|x|(in any case it would be positive so it is just a positive x here on out), r = irrational
n^r = -1•(x^r)
= (e^(i•pi))•(x^r)

Further,
Let (e^(i•pi))•(x^r) = y
-> e^(i•pi) = y / (x^r)
Ln both sides
-> i•pi = ln(y/(x^r))
-> 1 = i•(ln((e^(i•pi))•(x^r)) - ln(x^r)
-> 1 = i•( i•pi + ln(x^r) - ln(x^r))/pi
-> 1 = i•i•pi/pi
-> 1 = -1
Totally stumbled upon it by accident

[LwLiPp]

You can get the condition pretty fast with Cauchy's condensation test.

[zelda]

If we're ignoring the complex plane, then b can be 0. The limit of 0^0 exists in the reals; it's 1. It's only in the complex plane that the limit of x^x or sin(x)/x or similar function ratios don't exist. Is there a follow-up on this for the complex plane though? I'd like to see what the radius of convergence in the complex plane is constant or a function of theta.

[noellundstrom]

It would be interesting to see what happens if b is complex.

[mhamedchaim]

شكرا لك ،ارجو ان تنزل مقاطعا خاصة بالتكامل الثنائي خاصة كيفية تعيين الحدود

[patrickhodson]

My guess is |b|<e since normally |r|<1 for Σr^n. Idk though. I've completely forgotten how to do radius of convergence.

[jacks.]

I think you made a mistake at 10:38 . The inequality stays because e^x is an increasing function, not because ln x is an increasing function. The definition of increasing is "If a>b, then f(a)>f(b)$. We started with ln(b)<-1, and since e is increasing exp(ln(b)) < exp(-1).

[olivier]

Of b = e than this is just the sum of All natural Numbers which equals -1/12