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I know the Integral test is the following theorem:

Assume $f$ is continuous, positive, and decreasing on [$1, \infty$).

If $\int_1 ^{\infty}f(x)\,dx$ exists and is finite, then $\sum f(n)$ converges and vice versa.

I am searching for counterexamples to this test if:

(i) the condition positive is dropped;

(ii) the condition decreasing is dropped.

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  • $\begingroup$ My first thought is that counterexamples, while extant, may be a bit too pathological to define explicitly (I'm sure, however, that some clever analysis folks can find something). To me, it is more important to understand that these are sufficient conditions for $\int_1^\infty f(x)\,\mathrm{d}x$ to serve as a lower bound for $\sum_1^\infty s_n$ and as an upper bound for $\sum_2^\infty s_n$. tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx $\endgroup$ – Nico Nov 28 '16 at 4:11
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    $\begingroup$ If it is negative and decreasing, it can't even pass the test for divergence, right? $\endgroup$ – CPM Nov 28 '16 at 4:12
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$\sin^2(\pi n)$ converges as a summation, but not as an integral.

Edit: As Nico points out, you can make this strictly positive by adding $1/n^2$.

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    $\begingroup$ Which point is this addressing? If you are dropping the hypothesis of decreasing while keeping positivity, this doesn't work, while if you are trying to drop positivity while keeping decreasing, this also doesn't work. This function is zero at certain points, hence not positive there. $\endgroup$ – Alex Ortiz Nov 28 '16 at 4:20
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    $\begingroup$ I think you may be able to salvage it by adding $\frac{1}{n^2}$ to your example. $\endgroup$ – Nico Nov 28 '16 at 4:23
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    $\begingroup$ As pointed out in the comments above, negative and decreasing is trivial. This is meant to address why you cannot relax the decreasing condition. $\endgroup$ – T.J. Gaffney Nov 28 '16 at 4:28
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When you drop the "decreasing" condition, both directions fail even with the "positive" condition! $ \def\l{\left} \def\r{\right} \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\nn{\mathbb{N}} \def\rr{\mathbb{R}} $

$\sum_{n=1}^\infty \l( \sin(nπ)^2 + \lfrac1{n^2} \r) = \color{blue}{\lfrac{π^2}{6}}$ but $\int_1^\infty \l( \sin(nπ)^2 + \lfrac1{n^2} \r)\ dn = \color{red}{\infty}$.

$\sum_{n=1}^\infty \lfrac1{n^4 \sin(nπ)^2+1} = \color{red}{\infty}$ but $\int_1^\infty \lfrac1{n^4 \sin(nπ)^2+1}\ dn < \color{blue}{\lfrac{π^2}{6}}$.

To derive the above inequality, note the following for any $k \in \nn_+$ and $n \in [k,k+1]$:

  • $\int_k^{k+1} \lfrac1{n^4 \sin(nπ)^2+1}\ dn < \int_k^{k+1} \lfrac1{(k^4-1) \sin(nπ)^2+1}\ dn = \int_{-\frac12}^\frac12 \lfrac1{(k^4-1) \sin(xπ)^2+1}\ dx = \lfrac1{k^2}$.

    [To get the last integral, use the substitution $t = \tan(xπ)$.]

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  • $\begingroup$ In your integration how the limits of integration comes from $\displaystyle \int_k^{k+1}$ to $\displaystyle \int_{-1/2}^{1/2}$ ? $\endgroup$ – Empty Jan 12 '17 at 7:05
  • $\begingroup$ @S717717: Because $\sin(xπ)^2 = \frac12( 1 - \cos(2πx) )$ has period $1$ in $x$. $\endgroup$ – user21820 Jan 12 '17 at 7:34
  • $\begingroup$ User21820)) Ok. But I got the final answer of the integral is $\pi^2 /k^2$ instead of $1/k^2$. $\endgroup$ – Empty Jan 13 '17 at 16:09
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    $\begingroup$ @S717717: Um numerical integration agrees with me. Did you forget some factor somewhere? Letting $t = \tan(xπ)$ we get $\sin(xπ)^2 = \lfrac{t^2}{1+t^2}$ and $\lfrac{dt}{dx} = π \sec(xπ)^2 = π(1+t^2)$ and hence $\int \lfrac1{(k^4-1) \sin(xπ)^2 + 1} = \int \lfrac{1+t^2}{(k^4-1) t^2 + (1+t^2)}\ dx = \int \lfrac{1}{π(k^4 t^2 + 1)}\ dt = \cdots$. $\endgroup$ – user21820 Jan 13 '17 at 16:31
  • $\begingroup$ Okk..I got it..Sorry for my mistake..! $\endgroup$ – Empty Jan 13 '17 at 16:48
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There can't be a convergent series or integral that's negative and decreasing, since it would by definition be above a given absolute value for sufficiently large n or t. So for negative, decreasing, and continuous, it's trivially true.

For decreasing, there's no reason you can't have a continuous (albeit not differentiable) function like a saw, with lines running to $\frac{1}{x^2}$ at integers, and 2 at the halfway points between integers. That would be a counterexample, since the integral between any two positive integers would never drop below 1, so the integral wouldn't converge, even though the series would.

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Given a sequence $(f(n))_n$ such that $\sum_1^\infty f(n)$ converges, you can extend $f$ to a continuous real function that does just about anything you want on each $[n,n+1].$

For example for each $n,$ let $f(x)$ be linear on $[n,n+1/2]$ and linear on $[n+1/2, n]$ with $f(n+1/2)$ large enough that $\int_n^{n+1}f(x)\;dx>1.$ Then $\int_1^{\infty}f(x)\;dx=\infty.$ Of course $f$ is not monotonic.

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