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Integral Test November 22, 2008

Posted by putnam120 in Math Related.
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Well it’s been a while since I have posted a math related post. So I am going to do this one on one of the problems from our analysis homework.  Basically we were asked to prove the integral test, not too difficult but definitely something that should be done.

Statement: Assume that f(x)\ge 0 and that f decreases monotonically on \left[1,\infty\right). Then \displaystyle\int_1^\infty f(x)dx converges if and only if \displaystyle\sum_{n=1}^\infty f(n) converges.

Aside: When I submitted this to my professor for grading in proved the theorem in both direction. Here I am going to try and combine them, thus saving time on my part.

Proof: Consider the interval \left[m,n\right] where m,n are integers with m<n. Additionally let P be the partition \left\{m,m+1,\dots ,n\right\}. Now because f is monotonically decreasing and f\ge 0 we have

(1)   \displaystyle 0\le\sum_{k=m+1}^nf(k)=L(P,f)\le\sum_{k=m}^{n-1}f(k)=U(P,f)\le\sum_{k=m}^nf(k)

Let \epsilon>0. If \displaystyle\sum_{n=1}^\infty f(n) converges then there exists an $N$ such that \displaystyle\sum_{k=m}^n f(k)<\epsilon whenever m,n>N. Similarly if \int_1^\infty f(x)dx converges we have that there exists W such that \int_m^nf(x)dx<\epsilon whenever m,n>W. The theorem follows from combining these facts with (1). \mathbb{Q.E.D.}

I would like to mention that I have left out some of the small details, such as proving that the integral actually does exist if the sum converges.

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