Please help me decide if the following statement is true or false:
If $\int_{1}^\infty f(x)dx$ converges then $\lim_{x\to\infty}f(x)=0$
I tried many counter examples with no luck so I tried to prove it but couldn't pull it off either...
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Noam
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4This is not true, try to construct a function made up of triangles with area $\frac{1}{n^2}$, it does not have a limit, but its integral converges. (Like a saw where its teeth become thinner and thinner). – omega-stable Jul 10 '17 at 20:05
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See also. – Jul 10 '17 at 20:10
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See the link in the comments for counterexamples. However, if $\lim_{x\to\infty}f(x)$ exists then it is zero. To prove this let $$\lim_{x\to\infty}f(x)=l$$ and assume $l\neq 0$. Then for some $N$, $f(x)> l-|l|/2$ whenever $x\ge N$. We then have $$\int_1^{\infty}f(x)dx=\int_1^{N}f(x)dx+\int_N^{\infty}f(x)dx>\int_1^{N}f(x)dx+\int_{N}^{\infty}(l+\frac{|l|}{2})dx$$ however the RHS side of the inequality diverges, and thus so must the LHS. This then proves the statement by contrapositive.
Will Fisher
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