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Having trouble with limit problem

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the limit as n goes to infinity of [ 1/(n+1) + 1/(n+2) + ... + 1/(2n) ]

    2. Relevant equations



    3. The attempt at a solution

    I tried rewriting the sum so we can get the indeterminate form infinity/infinity and use L'Hopital's rule but the derivative gets too complicated. Is there some way to simplify the sum first?

    I asked my teacher for help but he said he can't help me with this.
     
  2. jcsd
  3. Sep 12, 2009 #2

    Mentallic

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    I don't quite understand your question.
    Since, isn't it obviously 0+0+...+0=0?

    I'm sure you mean something else :smile:
     
  4. Sep 12, 2009 #3
    My book says the answer is ln2. Are you sure it's 0?
     
  5. Sep 12, 2009 #4

    HallsofIvy

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    No, it is not. You take the sum for each n, then let n go to infinity. That is NOT the same as letting n go to infinity in each term. For example, when n= 2 the sum is 1/3+ 1/4= 7/12. When n= 3, 1/4+ 1/5+ 1/6= 74/120= 37/60. For general n the sum is, I believe,
    [tex]\frac{3n^2+ n}{2}\frac{(2n)!}{n!}[/tex]
    so the question is, what does that go to as n goes to infinity?
     
  6. Sep 12, 2009 #5

    VietDao29

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    Because the number of terms in that sum is n, which also tends to infinity as n increases without bound. Consider the following sum:

    [tex]\lim_{k \rightarrow \infty} \sum_{n = 1} ^ k \frac{1}{k} = \lim_{k \rightarrow \infty} \underbrace{ \left(\frac{1}{k} + \frac{1}{k} + ... + \frac{1}{k} \right)}_{k \mbox{ terms}}[/tex]

    Each term tends to 0, but the above limit is simply:

    [tex]\lim_{k \rightarrow \infty} \frac{k}{k} = 1 \neq 0[/tex].

    ----------------------------

    So, back to the OP's problem, when dealing with infinitely-many-term sum like this, one should make use of the definition of Riemann Integral. You can go, like:

    [tex]\sum_{i = n + 1} ^ {2n} \frac{1}{i}[/tex]

    [tex]= \sum_{i = n + 1} ^ {2n} \frac{1}{n} \left( \frac{n}{i} \right)[/tex]

    [tex]= \sum_{i = n + 1} ^ {2n} \frac{1}{n} \left( \frac{1}{\frac{i}{n}} \right)[/tex]

    By letting i = n + i', or similarly i' = i - n

    [tex]= \sum_{i' = 1} ^ {n} \frac{1}{n} \left( \frac{1}{\frac{n + i'}{n}} \right)[/tex]

    [tex]= \sum_{i' = 1} ^ {n} \frac{1}{n} \left( \frac{1}{1 + \frac{i'}{n}} \right)[/tex]

    Now, let's think of a (continuous) function, and a range, such that when applying Riemann Integral to integrate the function over that range will produce the above sum.

    ----------------------------

    I remembered reading a book, which has a nice approach on solving this problem that does not use integration. It's a little bit more complicated, you need to look at it quite closely to realize it. I'll show you in case it comes back to my mind again. :)
     
  7. Sep 12, 2009 #6
    I think the range will be [0,1/2] and the function will be decreasing on it's domain. I'm not seeing the explicit function of n.
     
  8. Sep 12, 2009 #7
    Write the sum as [itex]\sum_{k=1}^{n}\frac{1}{n+k} = \sum_{1}^{n}\left(\frac{1}{1 + \frac{k}{n}}\right)\cdot\frac{1}{n}.[/itex] Reinterpret this as the sum of the area of rectangles, where each rectangle has base [k/n, (k+1)/n] and height 1/(1 + k/n). Consider the function f defined by f(x) = 1/(1+x) on the interval [0,1]. Partition [0,1] into n equal subintervals by the points [itex]0 = t_0 < t_1 < \mbox{ ... } < t_n = 1.[/itex] Clearly, [itex]t_k = \frac{k}{n}[/itex] where k ranges from 0 to n. Since f is decreasing, the infimum of f over the interval [itex][t_{k-1}, t_k][/itex] is [itex]f(t_k)[/itex] and the supremum [itex]f(t_{k-1})[/itex]. The lower sum is [itex]\sum_{1}^{n}\left(\frac{1}{1 + \frac{k}{n}}\right)\cdot\frac{1}{n}[/itex] and the upper sum [itex]\sum_{1}^{n}\left(\frac{1}{1 + \frac{k-1}{n}}\right)\cdot\frac{1}{n}.[/itex] Since [itex]\sum_{1}^{n}\left(\frac{1}{1 + \frac{k}{n}}\right)\cdot\frac{1}{n}[/itex] is between the upper and lower sums, the area that will be assigned to f over [0,1] is equal to this last sum for arbitrarily large n, which is equal to the original limit. But this is just [itex]\int_{0}^{1}\frac{1}{x+1}\,dx,[/itex] which evaluates to ln(2).

    Note: This is a lot easier if you draw a picture of the function defined by 1/x or 1/(x+1). You may apply the squeeze theorem if you have the right geometric notion down.
     
    Last edited: Sep 12, 2009
  9. Sep 13, 2009 #8
    Wow thanks alot guys. I get it now but i am not sure how we know the limits of integration to be from 0 to 1. I can see how their difference is 1.

    VietDao29 do let me know of that other method also if you remember it.
    I really like this question :)
     
  10. Sep 13, 2009 #9

    VietDao29

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    Have you covered Riemann Sum in university yet? If you have, then you can have a brief review on that Chapter. Reading the concept once again, and scan through some examples may help. When reading examples, just keep asking question, like why did they do this. And try to find the answer for that question. You may also try to solve some of the problems about Riemann Sum in your text-book. It should be useful.

    Practicing solving problems is the key to spot things quickly. :)

    As snipez90's pointed out, the limit can be: [tex]\int_0 ^ 1 \frac{1}{1 + x} dx[/tex], and it can also be: [tex]\int_1 ^ 2 \frac{1}{x} dx[/tex], depending on how you partition, and which function you choose.

    Btw, we "often" partition some interval into equal sub-intervals.

    ---------------

    In case you haven't covered that in school, you can have a look here. There's an example at the bottom of the page: http://en.wikipedia.org/wiki/Riemann_sum" [Broken].

    Don't hesitate to shout it out, in case you find something unclear, the HHs, as well as other people here are more than willing to help you.

    Another method is here. Let's try to see if you can solve the problem using this method:

    https://www.physicsforums.com/showthread.php?t=336834

    (this is fairly hard, as to solve this, you'll need to prove some other limits (like, the Euler Constant, and stuff))

    But this is another way to go, so, just give it a shoot if you'd like. :)

    There's still at least one more way to go. I'll search through my books this afternoon, and hopefully, if I find it, I'll post it here.

    If I remember correctly, it involves some kind of exponential function. =.="
     
    Last edited by a moderator: May 4, 2017
  11. Sep 13, 2009 #10
    Ok so we let x = k/n. When k=1 and n-> infinity , x-> 0. And when k=n, x is just 1. This is why the limits of integration are 0 and 1.

    Thanks again guys, case solved :P.
     
  12. Sep 13, 2009 #11

    Mentallic

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    Aha! thanks for correcting me. It didn't cross my mind that the number of terms also tends to infinite.
     
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