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Math Help - Integral test

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Integral test

    Hello everyone! While doing some problems I came upon this one. What is disconcerting is that I have found (in general) that the more useful a theorem is the more difficult it is to prove. So I would appreciate if someone would tell me if this looks correct?

    Terminology:

    P_n a partition of [a,b] containing n points with x_1=a.

    \wp is the set of all partitions of [a,b]

    \Delta \alpha_i=\alpha(x_{i+1})-\alpha(x_{i-1})~~\alpha\in\uparrow

    M_i=\sup_{x_{i}\leqslant x\leqslant x_{i+1}}f(x)

    m_i=\inf_{x_{i}\leqslant x\leqslant x_{i+1}}f(x)

    Taking \alpha(x)=x

    U\left(P_n,f\right)=\sum_{i=1}^{n-1}M_i\cdot\Delta x_i

    L\left(P_n,f\right)=\sum_{i=1}^{n-1}m_i\cdot\Delta x_i

    Now suppose that f\in\mathcal{R} ( f is Riemann integrable) on [a,b] then \int_a^b f ~dx=\inf_{P_n\in\wp}U\left(P_n,f\right)=\sup_{P_n\  in\wp}L\left(P_n,f\right)

    Note: by how we defined \int_a^b f~dx it follows that for all partitions L\left(P_n,f\right)\leqslant \int_a^b f~dx\leqslant U\left(P_n,f\right)

    Ok now onto the question

    Question: Suppose that f(x) is a positive, monotonically decreasing function, prove that \sum_{x=1}^{\infty}f(x)\text{ converges}\Longleftrightarrow\int_1^{\infty}f~dx\t  ext{ converges}

    Answer: Part one \sum_{x=1}^{\infty}f(x)\text{ converges}\implies \int_1^{\infty} f~dx\text{ converges}. Consider the interval [1,b] with b\in\left\{2,3,4,\cdots\right\}. Define the P_n as being the set of n natural numbers in [1,b]. It is clear that \Delta x_i=1. Now consider the ith point in the partition. This point will be i (since we defined the partitions as the set of natruals). Then the interval [x_{i},x_{i+1}] will be the interval [i,i+1], so on any interval M_i=\sup_{x_{i}\leqslant x\leqslant x_{i+1}}f(x)=\sup_{i\leqslant x\leqslant i+1}f(x)=f(i) since f is monotonically decreasing. So U\left(P_n,f\right)=\sum_{i=1}^{n-1}M_i\cdot\Delta x_i=\sum_{i=1}^{n-1}f(i). So on any interval [1,b] we have that \int_1^b f~dx\leqslant\sum_{i=1}^{n-1}f(i). So now letting b\to\infty\implies P_n\to\mathbb{N} which in turn implies n\to\infty (since there are infinitely many elements of \mathbb{N}) so \int_1^{\infty} f~dx\leqslant\sum_{x=1}^{\infty}f(x) and since the RHS converges by the hypothesis this concludes the proof.



    Part two: \int_1^{\infty}f~dx\text{ converges}\implies\sum_{x=1}^{\infty}f(x)\text{ converges}. Define [1,b] and P_n as before, consqequently we have that \Delta x_i=1 again. Except this time m_i=\inf_{x_i\leqslant x\leqslant x_{i+1}}f(x)=\inf_{i\leqslant x\leqslant i+1}=f(i+1). So L\left(P_n,f\right)=\sum_{i=1}^{n-1}m_i\cdot\Delta x_i=\sum_{i=1}^{n-1}f(i+1). Now as was stated L\left(P_n,f\right)\leqslant \int_a^b f~dx so this implies \sum_{i=1}^{n-1}f(i+1)\leqslant\int_1^b f~dx. So once again letting b\to\infty\implies P_n\to\mathbb{N} which in turn implies n\to\infty so \sum_{x=1}^{\infty}f(x+1)\leqslant \int_1^{\infty} f~dx. Now since the LHS may be written as \sum_{x=1}^{\infty}f(x)-f(1) and the RHS converges this concludes the proof.

    Now combinging parts one and two gives \sum_{x=1}^{\infty}f(x)\text{ converge}\Longleftrightarrow\int_1^{\infty}f~dx\te  xt{ converges}\quad\blacksquare

    Hows that look?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    I have a question pertaining to this. Before I did it as such I considered just partitioning [1,\infty) by the naturals. Of course I realized that there were a plethora of problems with this but one in particular I wanted to ask about. I know that a partition P of [a,b] must have finite cardinality but does this still apply when either/or a,b is \infty,-\infty? I think the answer is yes, since technically \int_a^{\infty} f~d\alpha does not make sense in the Riemann-Stieltjes sense (we of course obviate this by writing \lim_{b\to\infty}\int_a^b f~d\alpha.
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