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Math Help - Infinite series related to ingral

  1. #1
    MHF Contributor arbolis's Avatar
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    Infinite series related to ingral

    Today I spent some time about an infinite series. I came up with \sum_{i=1}^{+\infty}\frac{1}{3^i}. I know it converges (thanks to the integral test of convergence. \int_1^{\infty}\frac{1}{3^x}dx=\frac{1}{3ln(3)}). Now I don't understand why the value of the integral is not the value of the series. A decreasing function f:[a,b]\rightarrow R has a definite integral \int_a^b f(x) dx equal to \frac{b-a}{n}\sum_{i=1}^n (t_i-t_{i-1})f(t_i). In our case f(x)=\frac{1}{3^x} is decreasing. Now if a=1, b and n tends to +\infty the integral becomes the infinite series I came up with, which is worth \frac{1}{2} and not \frac{1}{3ln(3)}. Where is my error? Also, if the integral I have and the infinite series I have are not equal, why can we use the integral test for convergence, while working on an infinite series? It would mean that if the series converges, then the integral does so, but their values are not equal... Strange. And of course, if an integral converges, so does its infinite series and their values are equal. But in my example I made an error, that's why it doesn't work.
    Last question : is it possible to determine to what value converges the series using integrals? Thanks in advance.
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    Because the integral test is only used to tell if a series converges or not. The series you are dealing with is a geometric series, you can easily sum it, why are you even using integral test for this? While the integral test gives you are different value.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Because the integral test is only used to tell if a series converges or not.
    ,
    While the integral test gives you are different value.
    . Yes I know that the integral test is used to determine if a series converges or not, I used it for this. I also know that the integral test gives me a different result, that's why I said
    It would mean that if the series converges, then the integral does so, but their values are not equal...
    . But I don't understand why the integral is not equal to the infinite series. The integral I gave is equal to an infinite series, but not the same of which I came up with. And I don't understand why not... Hope you understand me.
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    Because in the proof of the theorem that integral convergence is equivalent to series convergence it is never established that it is.
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    MHF Contributor arbolis's Avatar
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    Thanks, you made my mind. I just read the proof and understand better. Now my statement
    Now if , and tends to the integral becomes the infinite series I came up with
    is false, and I will take some time probably tomorrow (I didn't sleep last night) to see this properly. More precisely that \frac{b-a}{n}\sum_{i=1}^n (t_i-t_{i-1})f(t_i) is different from \int_1^{\infty}\frac{1}{3^x}dx when b and n tends to +\infty.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Today I spent some time about an infinite series. I came up with \sum_{i=1}^{+\infty}\frac{1}{3^i}. I know it converges (thanks to the integral test of convergence. \int_1^{\infty}\frac{1}{3^x}dx=\frac{1}{3ln(3)}). Now I don't understand why the value of the integral is not the value of the series. A decreasing function f:[a,b]\rightarrow R has a definite integral \int_a^b f(x) dx equal to \frac{b-a}{n}\sum_{i=1}^n (t_i-t_{i-1})f(t_i). In our case f(x)=\frac{1}{3^x} is decreasing. Now if a=1, b and n tends to +\infty the integral becomes the infinite series I came up with, which is worth \frac{1}{2} and not \frac{1}{3ln(3)}. Where is my error? Also, if the integral I have and the infinite series I have are not equal, why can we use the integral test for convergence, while working on an infinite series? It would mean that if the series converges, then the integral does so, but their values are not equal... Strange. And of course, if an integral converges, so does its infinite series and their values are equal. But in my example I made an error, that's why it doesn't work.
    Last question : is it possible to determine to what value converges the series using integrals? Thanks in advance.
    You cannot generally use an integral to find the actual value a series converges. But in certain cases as you said the infinite series is a Riemann sum that represents an integral

    For example

    \lim_{n\to\infty}\sum_{i=0}^{n}\bigg[\bigg(\frac{2i}{n}\bigg)^2\bigg]\frac{2}{n}=\int_0^{2}x^2+1dx

    As for your specific case since you have a sum of the form \sum_{n=0}^{\infty}x^n where |x|<1 we can say that that your series is equal to \frac{1}{1-x}

    \sum_{n=0}^{\infty}\frac{1}{3^n}=\sum_{n=0}^{\inft  y}\bigg(\frac{1}{3}\bigg)^n=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}
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