Example of a function for which $\int_1^\infty f(x)\;dx$ diverges and $\sum_{1}^\infty f(n)$ converges

Question

I recently came across this one:

A function which is continuous and positive for $x \geq 1$ and such that $\int_1^\infty f(x)\;dx$ diverges and $\sum_{1}^\infty f(n)$ converges

The author gives this answer:

I understand what the function $g$ is .

My questions are:

Is there anything special about the function $g$ here?
Why define $f$ by adding $g$ to $\frac{1}{x^2}$ ?
Why $f$ is continuous?
Is there any other simple functions $g$ instead of the given one to make $f$ easier?

For my second question, I think we know $\sum \frac{1}{n^2} < \infty $ and $g(n)=0$ for integer points. So we define like this. Is this correct?

For my third question, I think $g$ is continuous and $\frac{1}{x^2}$ is continuous, so their sum is continuous. Is there any reason apart from this?

Any help must be appreciated and thanks in advance!

Second bullet: you want $f$ to be positive, while $g$ is only nonnegative. — Clement C., Sep 27 '18 at 12:50

user · Accepted Answer · 2018-09-27T13:00:49.993

2

Let consider for example

$$g(x)=\sin^2(2\pi x) \implies g(n)=\sin^2 (2\pi n)=0\quad \forall n$$

The $1/x^2$ term has been added in order to have $f(n)>0$.

The continuity was considered in order to exclude simple example for $f$ as for example $f(x)=1$ $\forall x\not \in \mathbb{Z}$ and $f(x)=1/x^2$ otherwise.

edited Sep 27 '18 at 13:00

answered Sep 27 '18 at 12:50

user

154,566

It should be positive for $x\geq 1$ – Jakobian Sep 27 '18 at 12:53
@Jakobian Thanks, I lost that detail at first! – user Sep 27 '18 at 13:01
@gimusi: Actually we need $f(x) >0$ for all $x$. I'm misunderstand $f(n)>0$. can you explain it a bit more? – Chinnapparaj R Sep 27 '18 at 13:03
@ChinnapparajR In the OP I read positive for any $x\ge 1$ anyway the example given works for any $x\neq 0$ and if we use $1/(x^2+1)$ for any $x$. – user Sep 27 '18 at 13:06
@ChinnapparajR Without the term $1/n^2$ we would have $f(n)=0$ $\forall n$. – user Sep 27 '18 at 13:07
for all $x$ in my comment means for all $x$ in the domain of definition $[1,\infty)$ – Chinnapparaj R Sep 27 '18 at 13:10
@ChinnapparajR Then it ok $f(x)=\sin^2(2\pi x)+1/x^2>0$ for any $x\ge 1$. – user Sep 27 '18 at 13:11
So, In summary, you prove my fourth bullet, namely to make $f$ easier such as taking $g(x)=\sin ^2(2 \pi x)$, so the hypothesis and conclusions are remains true. Thank you! – Chinnapparaj R Sep 27 '18 at 13:16
@ChinnapparajR Yes this is a good choice for the $g(x)$ function. You are welcome! Bye – user Sep 27 '18 at 13:19

score 2 · Answer 2 · answered Sep 27 '18 at 13:11

2

As little constraints are imposed on $f$, you can very well choose a smooth function which is zero at integer values and positive elsewhere, such as $\sin^2(\pi x)$.

If the function must be strictly positive, add a term with sufficient decrease speed to guarantee convergence of the series, say $e^{-x}$ (the book chose $x^{-2}$).

answered Sep 27 '18 at 13:11

Thanks! This one really helps for the intuition of such functions – Chinnapparaj R Sep 27 '18 at 13:24

Example of a function for which $\int_1^\infty f(x)\;dx$ diverges and $\sum_{1}^\infty f(n)$ converges

2 Answers2

Linked