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I am wondering if there is a rigourous way to show the following series converges:

$$ f(x) = \begin{cases} \dfrac{1}{x^2} & x \in \mathbb{N} \\[1ex] g(x) & x \not\in \mathbb{N}\in \mathbb{R} \end{cases}$$

Here, $g(x)$ could be anything, but for interest, wouldn't converge on its own.

When we do the following: $$\lim_{k\to\infty}\sum_{n=1}^{k}f(n).$$

Do we get a limit that exists?

Intuitively we would, because summation is only defined on the natural numbers (unless we redefined it here, doing some interpolation between points). However, I am not certain this function is integrable, which would suggest the function does not converge (unless my understanding of the Integral Test is wrong, after all, it says nothing of whether the integral exists, just if it converges the sum converges and vice-versa).

rage_man
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    The notation $x \not\in\mathbb{N} \in \mathbb{R}$ is nonsense. Do you mean to say $x \in \mathbb{R}\setminus\mathbb{N}$? – Xander Henderson Jun 24 '20 at 14:26
  • The $\Sigma$ notation is usually used to iterate over integers. If so, the value of $f(x)$ outside $\mathbb{N}$ is irrelevant. – badjohn Jun 24 '20 at 14:28
  • After reading in more detail, I am confused about your question. Are you confused about the convergence of $\sum 1/x^2$? Or are you confused about the integral test? If the latter, please edit your question to indicate what you mean by "the integral test"---be as precise as possible, and do not leave out hypotheses. – Xander Henderson Jun 24 '20 at 14:38
  • @XanderHenderson What I am asking is does the given series converge. Re: notation, perhaps so, I wasn't sure of how to write what I meant to denote (a number outside the set of natural numbers). The expression does not seem to have an elementary antiderivative, or any integral at all, and so as a potential follow up I mentioned that that may impact some notion of the convergence of the function. – rage_man Jun 24 '20 at 14:44
  • @XanderHenderson Specifically, the confusion arises because the series is not well behaved in some sense of the word, and yet for the set of natural numbers it is, so does this allow for convergence? – rage_man Jun 24 '20 at 14:47
  • The series $\sum_{k=1}^{\infty} f(k)$ converges. There are many arguments here. – Xander Henderson Jun 24 '20 at 14:49
  • The summand of the series only needs to be defined, let along "well behaved", on the naturals. The business about $f(x) = g(x)$ for $x\not\in\mathbb{N}$ is a red herring. Even if your goal is to use the integral test, you need to be more careful---in order for the integral test to apply, $g$ cannot be arbitrary, Check the hypotheses. – Xander Henderson Jun 24 '20 at 14:52
  • Right -- if you were to add some variation of your first two sentences (about $g(x)$ being irrelevant), I would be able to accept that answer as it properly addresses the question. – rage_man Jun 24 '20 at 14:59

2 Answers2

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To address your confusion, we note that the integral test requires that $f(x)$ must be monotone decreasing over the interval $\left[1, \infty\right)$ to be applicable.

Check this link: https://en.wikipedia.org/wiki/Integral_test_for_convergence

Hence, if we have

$$f(n) = \frac{1}{n^2}$$

for all $n \in \mathbb{N}$, and you want to extend it to $\mathbb{R}$ by setting $f(x) = g(x)$ for all $x \in \mathbb{R} \setminus \mathbb{N}$, then you have to make sure that the selection of $g(x)$ keeps $f(x)$ monotone decreasing over the interval $\left[1, \infty\right)$ so that the integral test can kick in.

If the ultimate goal of extending $f(n)$ to the domain $\mathbb{R}$ as $f(x)$ is to hopefully construct $f(x)$ so that we can use the integral test to prove that the series

$$\sum_{n = 1}^{\infty} f(n) = \lim_{k \to \infty} \sum_{n = 1}^{k} f(n)$$

converges, then in this case, the way you extend $f(n)$ to $f(x)$ matters. If you can prove that there exists at least one way to extend $f(n)$ to $f(x)$ such that $f(x)$ is monotone decreasing over the interval $\left[1, \infty\right)$, then this is sufficient to invoke the integral test to prove that

$$\sum_{n = 1}^{\infty} f(n) < \infty$$

For instance, you can trivially take $g(x) = 1/x^2$ for $x \in \mathbb{R} \setminus \mathbb{N}$ so we have $f(x) = x^2$ for all $x \in \mathbb{R}$ which is monotone decreasing.

But if you just just want to know if the limit exists, then you can do so without extending $f(n)$ to $\mathbb{R}$. I mean, you can do so without invoking the integral test. Evaluating this sum is known as the Basel problem and the result is $\pi^2/6$.

Check this link: https://en.wikipedia.org/wiki/Basel_problem

Royce Pacibe
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About the series,

$$\sum_{n\ge 1} f(n)=\sum_{n\ge 1}\frac{1}{n^2}$$ converges as a Riemann series $(2>1)$.

As an integral, we can say nothing. It depends on the expression of $ f(x).$

Take for example $ f(x)=1$ if $ x\notin \Bbb N$, then

$$\int_1^{+\infty}f(x)dx \;\; diverges$$

and If $$(\forall x\ge 1)\;\; f(x)=\frac{1}{x^2}\; $$ the integral $$\int_1^{+\infty}f(x)dx\;\; converges$$

hamam_Abdallah
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    It is not entirely clear to me what the asker's question is: are they really interested in the series, or are they trying to understand the "integral test". Since the question is not entirely clear, it seems premature to provide an answer. – Xander Henderson Jun 24 '20 at 14:37