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:enter image description here

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!

  • 1
    $\begingroup$ Second bullet: you want $f$ to be positive, while $g$ is only nonnegative. $\endgroup$ – Clement C. Sep 27 '18 at 12:50

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.

  • $\begingroup$ It should be positive for $x\geq 1$ $\endgroup$ – Jakobian Sep 27 '18 at 12:53
  • $\begingroup$ @Jakobian Thanks, I lost that detail at first! $\endgroup$ – user Sep 27 '18 at 13:01
  • $\begingroup$ @gimusi: Actually we need $f(x) >0$ for all $x$. I'm misunderstand $f(n)>0$. can you explain it a bit more? $\endgroup$ – Chinnapparaj R Sep 27 '18 at 13:03
  • $\begingroup$ @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 $x$. $\endgroup$ – user Sep 27 '18 at 13:06
  • $\begingroup$ @ChinnapparajR Without the term $1/n^2$ we would have $f(n)=0$ $\forall n$. $\endgroup$ – user Sep 27 '18 at 13:07

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 $x^{-2}$).

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

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