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Here is part of Prob. 4, Chap. 7, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Consider $$ f(x) = \sum_{n=1}^\infty \frac{1}{1+n^2 x }. $$ For what values of $x$ does the series converge absolutely? . . .

My Attempt:

For $x = 0$, we have $$ \lim_{n \to \infty} \frac{1}{1+n^2 x} = 1\neq 0,$$ and so the series fails to converge, by Theorem 3.23 in Rudin.

For any $x > 0$, we see that $$0 < \frac{1}{1+n^2 x} < \frac{1}{n^2 x} = \frac{1}{x} \frac{1}{n^2},$$ and since $\sum \frac{1}{n^2}$ is convergent, so is our series, by Theorem 3.25 (a) in Rudin.

What about the values of $x < 0$?

And, what about complex values of $x$?

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  • $\begingroup$ It seem to me that for $x=-1$, the series is not defined. And not only for -1, but as well for countable number of values $x= \frac{1}{a^{2}}$, $a\in N$. Obviously for $x<-1$ you can use similar argument. It remians the question of $x\in (-1,0) - A$, where $A$ is the set where a term of a sum is not defined. $\endgroup$ – kolobokish Nov 27 '17 at 12:40
  • $\begingroup$ @kolobokish here is the link to another Math SE post of mine on this very problem in Rudin. math.stackexchange.com/questions/2539522/… Can you please have a look at that post too? $\endgroup$ – Saaqib Mahmood Nov 27 '17 at 13:06
  • $\begingroup$ @kolobokish here is the link to another Math SE post of mine on this very problem in Rudin. math.stackexchange.com/questions/2539522/… Can you please have a look at that post too? $\endgroup$ – Saaqib Mahmood Nov 27 '17 at 16:01
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For any $x\neq 0$ we have $|a_n|=\left|\dfrac 1{1+xn^2}\right|\sim \dfrac 1{|x|n^2}$ which is a term of a convergent series, so the initial series is absolutely convergent.

Of course this criteria operates for values $n\gg1$, but as kolobokish noticed, there is an issue when $x\in A=\{-\frac{1}{k^2}\mid k\in\mathbb N^*\}$.

If $x\notin A\cup\{0\}$ then $\sum\limits_{n=1}^{\infty}a_n$ is absolutely convergent.

For $x\in A$ then we can only speak about $\sum\limits_{n=n_0+1}^{\infty}a_n$ or $\sum\limits_{n=1\\n\neq n_0}^{\infty}a_n$ for $n_0=\sqrt{-\frac 1x}$.


what is the meaning of the symbol ∼ in your answer? Can you please elaborate?

This symbol means asymptotically equivalent : https://en.wikipedia.org/wiki/Asymptotic_analysis

$f(x)\sim g(x)\iff \lim\limits_{x\to\infty}\dfrac{f(x)}{g(x)}=1$ or here with sequences $\lim\limits_{n\to\infty}\dfrac{a_n}{b_n}=1$.

There is this theorem for comparing series:

  • Let $\sum a_n$ and $\sum b_n$ be two series with positive terms
  • If $a_n\sim b_n$ then $\sum b_n$ converges $\iff\sum a_n$ converges


By reversing sign, it also works for series with only negative terms, in our case since we study $|a_n|$, so the question of sign is trivially verified.

When we have a series with a complicated term $a_n$ we try to reduce it to a known convergent series, here $\sum\frac 1{n^2}$ by noticing $a_n=\dfrac 1{n^2x}\times\underbrace{\dfrac 1{1+\underbrace{\frac 1{n^2x}}_{\to 0}}}_{\to 1}$ so $a_n\sim \frac 1{n^2x}$.

Search your book, I'm pretty sure this theorem is there somewhere, maybe next chapter.

Although this come for the fact that if sequences are equivalent (i.e. $\frac {a_n}{b_n}\to 1$), one can find for $n\ge n_0\gg 1$ : $c_1 b_n\le a_n \le c_2 b_n$

And the partial sums verify the same inequalities $c_1\sum\limits_{n=n_0}^Nb_n\le\sum\limits_{n=n_0}^Na_n\le c_2\sum\limits_{n=n_0}^Nb_n$

Thus the series are of same nature (remember series with positive terms are $\nearrow$).

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    $\begingroup$ what is the meaning of the symbol $\sim$ in your answer? Can you please elaborate? $\endgroup$ – Saaqib Mahmood Nov 27 '17 at 16:04
  • $\begingroup$ edited to answer your question. $\endgroup$ – zwim Nov 27 '17 at 17:18

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