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[prove] Divergence of series

  1. Nov 27, 2008 #1
    Hello! I got one issue with proving divergence of series. I start covering this part of mathematics and don't understand how to prove it. Here is the issue:
    I got one harmonic series:

    [tex]\sum_{n=1}^{\infty}{\frac{1}{n}}=1 + \frac{1}{2} + \frac{1}{3} +...[/tex]
    We need to show that the series of partial sums (separate sums) is not bounded.

    Xn=1 + 1/2 +1/3 +...+ 1/n

    As I can see:

    X2=1 + 1/2 = X1 + 1/2

    but what I can't understand is:

    X4=X22=1 + 1/2 + 1/3 + 1/4 > 1 + 1/2 + 2*1/4 = 1 + 2/2

    and

    X2k>1 + k/2 where k>1

    Can you please give me a short explanation that would help me understand?

    Thanks in advance.
     
  2. jcsd
  3. Nov 27, 2008 #2

    Office_Shredder

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    Basically you group up the terms, doubling the number of terms grouped each time. So you start off

    (1) - 1 term. Now we add two more

    (1) + (1/2 + 1/3) - now we add four more

    (1) + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) - now we add 8 more

    (1) + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + (1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15)

    note the kth block has as its last term 1/(2k-1)

    Basically, each of those blocks of terms is greater than 1/2. Why?

    1 > 1/2 trivially

    1/2 + 1/3 > 1/2 trivially too.

    1/4 + 1/5 + 1/6 + 1/7... well, 1/4>1/8, 1/5>1/8, 1/6>1/8, 1/7>1/8... so

    1/4 + 1/5 + 1/6 + 1/7 > 1/8 + 1/8 + 1/8 + 1/8 =1/2

    Similarly for the next block

    1/8>1/16, 1/9>1/16.... 1/15>1/16 so

    (1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15) > 8/16 = 1/2

    You can repeat this process indefinitely.... the next 16 terms will all be greater than 1/32 ( so adding them up gives something > 16/32), the next 32 terms after that will all be greater than 1/64, so adding them gives something > 32/64, etc.
     
  4. Nov 27, 2008 #3
    Thanks for the post. It is great.

    But how do they got this one:

    X4=X22=1 + 1/2 + 1/3 + 1/4 > 1 + 1/2 + 2*1/4 = 1 + 2/2
    We said that 1 + 1/2 > 1/2, so 1+1/2+1/3+1/4>1/2+1/3+1/4. But it does not match up with the statement above. Can you help me realize it?

    Thanks in advance.
     
  5. Nov 27, 2008 #4

    mathman

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    It is based on the fact that 1/3 > 1/4 and for the next group 1/5, 1/6, 1/7 are all > 1/8, etc.
     
  6. Nov 27, 2008 #5
    Alternatively, you can use the integral test. It's easy to see geometrically. 1/n is the area of the rectangle with height 1/n and width 1 on the interval [n, n+1]. Looking at 1, 1/2, 1/3, 1/4, 1/n you can see that each rectangle is over the portion of the region under the curve 1/x delimited by x = n and x = n +1. So the area of the region under 1/x over the interval [1,n+1] is clearly smaller than 1 + 1/2 + ... + 1/n. That area is log (n+1), and that goes to infinity for large n's, so it follows that 1 + 1/2 + ... + 1/n also goes to infinity.
     
  7. Nov 28, 2008 #6
    @Werg22, sorry but I have never learned about integrals. In my next lessons I would probably do it.

    @mathman
    1+1/2>1/2
    and 1/3>1/4
    If we sum all of them + 1/4
    So 1+1/2+1/3+1/4>1/2+1/4+1/4=1/2+1/2=2/2
    Where is the 1 ?
     
  8. Nov 28, 2008 #7

    Office_Shredder

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    You tell me where the 1 is. You shouldn't have used the inequalities so bluntly

    1=1
    1/2=1/2
    1/3 + 1/4 > 1/2

    Hence 1 + 1/2 + 1/3 + 1/4 > 1 + 1/2 + 1/4 + 1/4 = 1 + 1/2 + 1/2 = 1 + 1 = 2
     
  9. Nov 28, 2008 #8
    Sorry for misunderstanding. Now I think I got it right:

    1/3 > 1/4

    1 + 1/2 + 1/3>1 + 1/2+ 1/4

    1+1/2+1/3+1/4>1 + 1/2 + 1/4 + 1/4

    So 2n>1 + n/2

    Thanks for the help.
     
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