1

If I have a factor inside a series that is bounded $\forall n \in \mathbb{R}$ (but this should apply also if it is definitely bounded, so for $n \to +\infty$) can I just study the convergence/divergence of the series without the factor?

For example, if I have to study the absolute convergence of the series $$ \sum_{n=1}^{\infty}\sin\left(\frac{1}{x^\alpha}\right)(\sqrt[n]{n} - \sqrt[n+1]{n}) \text{, } \alpha \in \mathbb{R} $$ I can just say that $\left|\sin\left(\frac{1}{x^\alpha}\right)\right|$ is bounded to $[0,1]$ and so $$ \sum_{n=1}^{\infty}\left|\sin\left(\frac{1}{x^\alpha}\right)\right|(\sqrt[n]{n} - \sqrt[n+1]{n}) < \sum_{n=1}^{\infty}\sqrt[n]{n} - \sqrt[n+1]{n} $$ but can I do the same for $ \sum_{n=1}^{\infty}\left(1 - \cos\left(\frac{1}{x^\alpha}\right)\right)(\sqrt[n]{n} - \sqrt[n+1]{n})$? Because $1 - \cos\left(\frac{1}{x^\alpha}\right)$ is bounded to $[0,2]$ instead!

One way I could motivate this is $$ \sum_{n=1}^{\infty}\left(1 - \cos\left(\frac{1}{x^\alpha}\right)\right)(\sqrt[n]{n} - \sqrt[n+1]{n}) \text{ converges} \iff \sum_{n=1}^{\infty}\frac{1 - \cos\left(\frac{1}{x^\alpha}\right)}{2}(\sqrt[n]{n} - \sqrt[n+1]{n}) \text{ converges} $$ now $\frac{1 - \cos\left(\frac{1}{x^\alpha}\right)}{2}$ is bounded to $[0,1]$, so I can say that $$ \sum_{n=1}^{\infty}\frac{1 - \cos\left(\frac{1}{x^\alpha}\right)}{2}(\sqrt[n]{n} - \sqrt[n+1]{n}) < \sum_{n=1}^{\infty}\sqrt[n]{n} - \sqrt[n+1]{n} $$ but is this a consistent method?

Mattia Gregnanin
  • 43
  • 4
  • Food for thought: the constant function $0$ is bounded, and if you multiply it to the terms of a series, then the result is always convergent, regardless of the convergence of the original series. You might also appreciate this question. – Theo Bendit Nov 18 '21 at 19:33
  • @TheoBendit and it is true that $\sum_{n=1}^{\infty}{0 \cdot a_n}$ converges when $\sum_{n=1}^{\infty}{a_n}$ does. I just cannot say anything when $\sum_{n=1}^{\infty}{a_n}$ diverges. Or am I wrong? – Mattia Gregnanin Nov 18 '21 at 19:39
  • You are right. The linked question helps show the other direction: when $\sum a_n$ converges and $b_n$ is bounded, then $\sum a_n b_n$ converges too. Note, this gives you the other direction if you can also show that $b_n^{-1}$ is bounded. – Theo Bendit Nov 18 '21 at 19:47
  • @TheoBendit thanks! So, for example, I cannot say that $\sum_{n=1}^{\infty}{\sin(n)a_n}$ diverges when $\sum_{n=1}^{\infty}{a_n}$ does because $\csc(x)$ is not a bounded function. But if I have $\sum_{n=1}^{\infty}{(2 + \sin(x))a_n}$ I can, right? – Mattia Gregnanin Nov 18 '21 at 20:00
  • Yes, that's right (excusing the $x$ that should be an $n$). – Theo Bendit Nov 18 '21 at 20:28
  • @TheoBendit yes, I was considering more generally $\csc : \mathbb{R} \to \mathbb{R}$. Thanks! – Mattia Gregnanin Nov 18 '21 at 20:44

0 Answers0