Solution: A Cool Double Summation + Integral Combo

  • Published on Aug 16, 2018
  • We explore four different methods of simplifying the expression to unity: integration by parts/reduction, Laplace transform, gamma function, and switching summation and integral.
    Congratulations to Mohamed S. Fawzy, Benjamin Wang, Hung Hin Sun, Gabriel N., Allan Lago, GreenMeansGO, Phoenix Fire, Diego Viveros, Cesare Angeli, and Foster Sabatino for successfully solving this week's math challenge question! Mohamed S. Fawzy was the first person to solve the question.
    Your support is truly a huge encouragement.
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Comments • 43

  • LetsSolveMathProblems
    LetsSolveMathProblems  11 months ago +14

    *At 15:21, I should have written +1/a*e^(-ax), NOT -1/a*e^(-ax). I apologize for the error; the final answer is correct as it stands. Thank you so much, GreenMeansGO, for notifying me! =)

  • Ethan Bartiromo
    Ethan Bartiromo Month ago

    400th like

  • MountainC
    MountainC Month ago

    For some reason I keep getting -1 instead of +1 for the answer.
    I believe that my steps are correct:
    interchange b_sum and integral with uniform convergence;
    get [ e^(-x(a-1))-e^(-ax) ] for the integrand;
    Integrate and get [ -e^(-x(a-1))/(a-1) + e^(-ax)/a ] from 0 to +infinity;
    Since the exponent goes to negative infinity as X gets bigger, all the terms involving infinity are zero;
    The summand becomes [ -1/(a-1) + 1)a ];
    Rearranging gives [ 1/a - 1/(a-1) ];
    Split up the sum and get the harmonic series from 2 minus the harmonic series from 1;
    Finally take out the first term of the sum from 1, i.e. (-1), and see that the remaining sums cancel out.
    Thus I got -1 for the answer.
    Can someone point out any flaws I made in this argument?

    • MountainC
      MountainC Month ago

      Get a piece of paper and try doing it to see what I mean

  • Pranjal Jaiswal
    Pranjal Jaiswal 2 months ago

    The laplace approach was very clever

  • Ronald Ssebadduka
    Ronald Ssebadduka 2 months ago

    Bro in integration you increase the powers not reduce them...what you did was differentiate in the first method of integration by parts.

  • Kutluhan Bilgeşah Bayzan

    İ dont looked solution My solution is equal to 1

  • Vermeer Dragyl
    Vermeer Dragyl 3 months ago +2

    One of a VERY FEW problems that I could solve. I love it. I didn't know Laplace and Gamma will play their part (Not that I know about their existences before, Lol). Thank you.

  • Lucas Hoffses
    Lucas Hoffses 5 months ago

    3:40 I smell gamma funcion. I’m sad you didn’t use that.

    • Agfd
      Agfd 3 months ago

      he used it later

  • Sanjitha R
    Sanjitha R 6 months ago

    very interesting problem..very well explained..

  • Tekaruchi Nodotome
    Tekaruchi Nodotome 6 months ago +3

    The definition of beautiful calculus problem

  • George Marklin
    George Marklin 10 months ago

    Simple method: Do b sum first. It's just a Taylor series for exponential minus first term
    sum(b=1,Inf)(x^b/b!)=(e^x-1) Then do the "a" sum. It's just a geometric series starting at quadratic term
    sum(a=2,Inf)(e^(-x))^a=(e^(-x))^2/(1-e^(-x)) Then simplify to get Integral(0,Inf)(e^(-x))dx=1

  • Wedemeyer Math
    Wedemeyer Math 10 months ago

    Cool problem. Thanks for sharing :)

  • Brian Connelly
    Brian Connelly 10 months ago +1

    What do you use to capture your writing?

  • Susie Hue
    Susie Hue 11 months ago

    Doesn't the sequence also have to be uniformly convergent for us to be able to switch the summation and integral?

  • Frenly Espino
    Frenly Espino 11 months ago


  • Miyuki Umeki
    Miyuki Umeki 11 months ago


  • Matthias Schulte
    Matthias Schulte 11 months ago

    A very nice problem and a decent solution, although I have to admit that I have forgotten that the Taylor series of e^x starts at 0 at first. ^^ Great video, keep going! :D

  • Math Machine
    Math Machine 11 months ago

    Did the math, calling it now, 1

  • Minecraftster148790
    Minecraftster148790 11 months ago

    11:45 you could also realise that it is the reciprocal of the formula for the n-1th triangle number, but divided by 2. The sum of the reciprocal of triangle numbers is 2, and when divided by 2 it equals 1

  • Jean le Ronde d'Amelbert
    Jean le Ronde d'Amelbert 11 months ago +2

    Instead of moving the integral outside of the first sum, isn’t it easier to move it outside both sums then evaluate the sums (they split into two sums multiplied) and THEN take the integral. I suppose he did not show this method for brevity.

  • Avana
    Avana 11 months ago +18

    I regret not trying this one, it looked way too hard so I just supposed that it would be way above my level of knowledge, but after seeing the solution, I have now concluded that with a bit of hard work, I could have definitely solved this. God damnit.

  • rakumanu
    rakumanu 11 months ago

    Fun fact: the last series is a half of the sum of the reciprocal of the triangular numbers.
    sum_(a=2)^infinity 1/((a-1)*a) = 1*1/2 + 1/2*1/3 + 1/3*1/4 + 1/4*1/5 + ... =
    = 1/2 * (1 + 1/3 + 1/6 + 1/10 + 1/15 + ...)
    which, of course, is equal to 1/2 * 2 = 1

  • Adlet Irlanuly
    Adlet Irlanuly 11 months ago +6

    Awesome!!! One of the most favorite channels.

  • Felipe Lorenzzon
    Felipe Lorenzzon 11 months ago +9

    I have never understood why on earth the Gamma function is more common than the Pi function, which is shifted to coincide with the factorial for integers

  • Joel Courtheyn
    Joel Courtheyn 11 months ago +5

    This is really a jewel !

  • Ranjan Bhat
    Ranjan Bhat 11 months ago +3

    Wow I can't believe you made a vid on this problem. I had just found this problem least week and I solved it in the third method and I feel so smart rn

    • LetsSolveMathProblems
      LetsSolveMathProblems  11 months ago +1

      When I made this problem, I did not think of using gamma function to simplify the integral. I have to thank you and other clever commenters who discovered the shortcut. =)

  • Snuzbei
    Snuzbei 11 months ago

    Nice solution, but wouldn't it be quicker to simplify the b summand using the Taylor series expansion for exp(x) about 0, and simply apply an integration to be left with the telescoping sum as required?

    • LetsSolveMathProblems
      LetsSolveMathProblems  11 months ago

      I believe I outlined your solution at 14:20. Thank you for your compliment! =)

  • Abbas Kagdi
    Abbas Kagdi 11 months ago

    15:30 mistake in partial fraction

  • Martin Epstein
    Martin Epstein 11 months ago +3

    Somehow I didn't get a lot of practice with telescoping sums and I always forget to look for them. I evaluated that last summation by recognizing the coefficients of the Taylor series of 1/(1-x) integrated twice.

  • Plutarch Heavensbee
    Plutarch Heavensbee 11 months ago +1

    I forgot about partial fraction decomposition

  • San Seng
    San Seng 11 months ago +2


  • Jared Jonas
    Jared Jonas 11 months ago

    This one was really cool! I solved it by making the integrand look like a gamma distribution PDF, which when integrated over the whole domain, equals 1 (after taking out a constant). That made the inside simplify to (1/a)^(b+1), which is essentially just a geometric series without the first two terms. That then made the outermost series a telescoping series, and in the end, you get 2-1 :)

  • GreenMeansGO
    GreenMeansGO 11 months ago +11

    15:21 should be +1/a*e^(-ax) not minus.

  • Ben Burdick
    Ben Burdick 11 months ago +21

    I'm really mad at myself for not trying this one! This is definitely something a calc 2 student could solve!

    • Stanley Nicholson
      Stanley Nicholson 11 months ago +1

      Ben Burdick same here, this is a great integral for calc 2 students

  • Miguel B.
    Miguel B. 11 months ago +4

    really nice video!

  • Andy Arteaga
    Andy Arteaga 11 months ago +5

    Soooooo cool!