An Infinite Sum from the Berkeley Math Tournament

[kornelviktor]

Dont worry if blackpenredpen retires we have this guy to teach us calculus

[afefd]

Here from bprp, amazing video! You went through everything with full clarity.

[rogiertp]

you are such a chill guy to listen to, also you made sure everyone that wasn’t comfortable with reindexing factorials got comfortable

[diddykong]

Simpler solution: n.n.n = n.(n-1).(n-2) +3.n.n -2.n = n.(n-1).(n-2) +3.n.(n-1) +n; each of these is zero for some early terms in the sum, that we can duly drop; in the remaining terms, each of our three terms cancels out the first factors of the n! denominator, leaving a sum from n = k to infinity of 1/(n-k)!, for k in {0, 1, 2} with a factor of 3 on the k = 2 one; we can re-index each of these to m = n-k to make it a sum from m = 0 to infinity of 1/m!, so we have e +3.e +e = 5.e.

[srevere]

Cool solution! Another way you can approach the problem is to see that the summand (n^3/n!) is already very similar to the expansion for e^x, but with n^3 instead of x^n. After some playing around, you might notice that if you plug in "e^x" itself into e^x, you get: e^(e^x) = sum( (e^x)^n / n! ), and if you take the third derivative, you get sum( (n^3 / n!) * (e^nx) ). Evaluating this at x = 0 is precisely the sum. So the solution is the third derivative of e^(e^x) evaluated at 0, which if you compute it, is 5e. And this approach generalizes to higher powers of n in the numerator.
Do keep in mind though that this problem was given as the third problem on the calc tiebreaker test, meaning that the students had only 15 minutes to do ALL three problems on the test. The approach above (or similar) might be more efficient or even the only option in those kinds of conditions.

[mdperpe]

The series equals (x (d/dx))^3 e^x evaluated at x=1. The left hand side expands to x e^x + 3x e^x + x^2 e^x. Setting x=1 gives 5e.

[AryanChhotray]

This is so underrated the explanation was so clear, you just earned another sub

[jacoboribilik]

If you've done some probability you recognise by multiplying by e^-1 and e that you have E(N^3), the third moment of a Poisson distributed random variable. By differentiating the Moment Generating function three times and evaluating at 0 you solve the problem.

[HL-iwdu]

If you wanted to generalize this to n^k/n!, I believe OEIS sequence A000110 gives you the coefficients of e. This sequence is the Bell numbers, or the number of ways to partition a set of exactly k elements. Correct me if I’m wrong.

[Mathguy]

Nice approach, but it is actually a lot easier to use the series expansion of eˣ.
eˣ = Σₙ xⁿ/n!

Differentiate and multiply by x three times, you will get
x(1+3x+x²)eˣ = Σₙ n³xⁿ/n!

Plug in x=1, final result:
5e = Σₙ n³/n!

[doomcake]

This was excellent! I appreciate you explicitly explaining what you did at 11:30 and again at end. Looking forward to your next video!

[spaghetti]

I did similar approach to Tony Wang (the guy in the video) but in the most efficient way in a few minutes. It essentially boils downs to the following decomposition of the numerator:
n^3
= n(n^2)
= n[(n-1)^2+2n-1]
= n[(n-1)((n-2)+1)+2(n-1)+1]
= n(n-1)(n-2) + 3n(n-1) + n.

Those terms become e+3e+e. I did this while reindexing between each step and replacing the 1/n! terms with e. The reindexing doesn't make any rational constants appear because each time before you cancel with the factorial, you remove the first term which is always zero (so index goes up 1, reindex brings it back down 1). At least for me, this seems to be the most direct and quick calculation.

Without the reindexing, my approach can be interpreted as writing n^3 as the sum of falling factorials which are like the power functions (x^n) of discrete calculus. If one has a background in combinatorics, the coffiencients are the third row of the Stirling numbers of the second kind: 1, 3, 1. Each falling factorial when sum with /n! leaves the sum unaffected (for reasons already discussed), so e. The sum of the kth row of Stirling numbers of the 2nd kind is the kth Bell number. So for n^k instead n^3, the result would be e*B_k. This is pretty well known so a combinatorist would know the answer at a glance.

[Ricocossa]

I did it much more simply. You just have to rearrange:

n^3 = n(n-1)(n-2) + 3n(n-1) + n.

Then you see immediately that the answer is (1+3+1)e = 5e.

[yashmehta]

A much shorter (and easier to motivate) solution is to simply write n^3 = n(n-1)(n-2) + 3n(n-1) + n, to make it easily cancel out with the factorials. Now we simply have 5 terms (1+3+1) with the same sum e, giving 5e.

[Bodyknock]

Nice problem. 🙂 The only thing I’d do slightly differently is simply reindex some of those sums so the summands are using n! as a factor rather than, say, (n-2)! Or (n-3)! . You end up in the same place but it’s a bit more streamlined.

[ccolombe]

You can actually use a trick I learned from a physics textbook way back in the day.

Notice that: n * x^n = x (d/dx x^n). You can repeatedly apply this operation to get:
sum (n^3 x^n / n!) = x (d/dx (x d/dx ( x d/dx (sum x^n /n!)))) .
The sum in the inner parens is e^x and then you just take derivatives and find sum(n^3 x^n) = x(e^x x (x + 2) + (x+1) e^x).

Evaluate that bad boy at x = 1 and you have your 5e. This trick can be used in all kinds of sum problems!

[Wagon_Lord]

Interesting. I solved it in a much easier way using telescoping sums.

The sum of

(n+1)^3/(n+1)! - n^3/n! will be zero.

But each term is equal to ((n+1)^2-n^3)/n!, or (n^2+2n+1-n^3)/n!

We can then repeat the above, changing the power to 2, then to 1 (the infinite telescoping sum will always be zero).

When we make the power 1, we get that one of the terms is the sum of 1/n!, which we know is e, and since it telescopes to zero, the other sum, n/n! is also e (which can be shown in an easier fashion)

In the power of 2 case, we have sum[(n+1-n^2)/n!]=0, or in other words, 2e - sum(n^2/n!)=0, so the n^2 sum is 2e

In the power of 3 case, we therefore have 2e+2e+e-sum(n^3/n!)=0, so the sum must be 5e

This method gives us an easier way to evaluate higher powers in the sum, for example

sum[(n+1)^4/(n+1)! - n^4/n! ]=0
sum[(n^3+3n^2+3n+1- n^4)/n! ]=0
5e+6e+3e+e-sum(n^4/n!)=0
sum(n^4/n!)=15e

We can also see sum(n^5/n!)=52e

I did a few more terms in the sequence, OEIS suggests these coefficients give the "Bell or exponential numbers: number of ways to partition a set of n labeled elements."

[pineapplejoosh]

Awesome explanation, I could understand every step! 👍🏻

[seendidthegreat]

Son- Mom? Can we watch bprp at the theatres? Mom- No baby, we have bprp at home! -bprp at home- this video :) all jokes aside thank you Tony. I appreciated this video because it brought back nice year one calc memories and makes me look forward to calc 3!

[Kevin-ygsm]

Loved your video, hoping to see more .