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A weird convergent series involving integral

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Let [tex]y_{n} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} - \int_{1}^{n} \frac{1}{t}dt[/tex] Prove that the sequence [tex] \{y_{n}\}_{i=1}^{n} [/tex] converges


    2. Relevant equations



    3. The attempt at a solution
    [tex] y_{1} = 1 [/tex]
    [tex] y_{2} = 1 +\frac{1}{2} - \int_{1}^{2} \frac{1}{t}dt = 1 + \frac{1}{2} - ln2 [/tex]
    [tex] y_{3} = 1 +\frac{1}{2} + \frac{1}{3} - \int_{1}^{3} \frac{1}{t}dt = \frac{11}{6} - ln3 [/tex]

    Now we can see that in a base case y_n is decreasing. If I can finish induction (and then show it is bounded, I can show it is convergent.

    So then Assume y_n < y_n-1 to show y_n+1 < y_n.
    I get a really nasty inequality that I am going to try to type in, and this is where I get stuck on this problem.

    y_n+1 = 1 + 1/2 + ... + 1/n + 1/(n+1) - [ [tex] \int_{1}^{n} \frac{1}{t}dt + \int_{n}^{n+1} \frac{1}{t} dt[/tex] ]

    = 1+1/2+...+1/n+1/(n+1) - ln(n) - ln(n+1) + ln(n)
    < [1+ 1/2 +...+ 1/n - ln(n)] + 1/(n+1) - ln(n+1) + ln(n) and by the induction hypothesis we get
    < [1 + 1/2 + ... + 1/(n-1) - ln(n-1)] + 1/(n+1) - ln(n+1) + ln(n) and obviously
    < 1+ 1/2 + ... + 1/(n-1) + 1/n - ln(n-1) - ln(n+1) + ln(n)
    = 1 + 1/2 + ... + 1/n - ln(n+1) %comment(ln((n^2+n)/n) = ln(n+1))
    < 1 + 1/2 + ... + 1/n - ln(n) = y_n

    So I get what I wanted, and I don't think I made any mistakes, but this is about as ugly as it gets. Does anyone have any more elegant ways to do what my goal is (see way up there some where showing {y_n} converges
     
    Last edited: Nov 12, 2008
  2. jcsd
  3. Nov 12, 2008 #2

    Office_Shredder

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    Are you trying to prove that yn converges? Draw a graph of 1/t, and then partition the x axis on the natural numbers and draw the upper and lower step functions of 1/t on that partition . That should give you a pretty good idea on some bounds of how large yn is
     
  4. Nov 12, 2008 #3
    Yes, I am trying to prove {y_n} converges. I see 1/t, and geometrically it is the area under the curve. But I think you just made me think of something else. We know that 1/t is integrable on a certain interval from 1 to n. Also, if it is integrable then the lower integral = upper integral = the integral. I wonder if I express the integral as one of these expressed as a sum, if I would get enough cancellations, and see a finite limit for the sequence. I'll go try that. Thank you!
     
  5. Nov 12, 2008 #4

    Dick

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    The sum of the series part is greater than an upper sum for the integral part. The extra part, the 1/n is not part of the upper sum but that's ok, it's still greater. That tells you that series minus integral is always positive. So y_n is bounded below by 0. Now go back to your idea to show y_n is decreasing. Can you prove that? If y_n is bounded below and decreasing, then it converges, right? You are not going to be able to get an exact limit by getting cancellations.
     
  6. Nov 13, 2008 #5
    First, I am scared to pick an upper sum of that integral. Because it is decreasing the sum would be
    [tex]\sum_{i=1}^{n} \frac{1}{x_{i - 1}}*(x_i - x_{i-1})[/tex] and the first term is 1/0. I know I am doing something wrong here but I don't know what?

    Yes, thank you. To show it is decreasing, I am going to pick an upper sum, since it is guaranteed to be less than or equal to the integral and use this to make a y'_n. Going with I am wrong then y'_n is going to have a general term of either 1/n or (1 + 1/n) depending on why my upper sum is wrong. Either way, this is easy to show decreasing. And since y_n is less than or equal to y'_n it too must be decreasing.Then we have a really nice theorem that every bounded monotonic sequence is convergent. So I guess I just have to work out what I am missing on the upper sum of the integral given here.
     
    Last edited: Nov 13, 2008
  7. Nov 13, 2008 #6

    Dick

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    Take y_3. An upper sum for the integral of 1/t from 1 to 3 is just 1+1/2. There is no 1/0 to be afraid of. Is there? The integral starts at 1, not 0.
     
  8. Nov 13, 2008 #7
    Yes I see. I had it backwards. Taking the upper sum of a decreasing function means that I must be taking f(x_i) where x_i is the leftmost point in some interval, not x_(i-1) like I was thinking. Of course, once I actually drew a (correct) picture this was easy to see. Thanks again.
     
  9. Nov 13, 2008 #8

    Dick

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  10. Nov 13, 2008 #9
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