0
$\begingroup$

The way the book proved it is this:

Its idea was to prove that the sequence is Cauchy then it follows that it is convergent in $\mathbb{R}$ ; enter image description here

But my way that I want to check if it's right or wrong is :

$$ \begin{aligned} \left|U_{n+p}-U_{n}\right| &=\left|\frac{\cos 5^{n+1}}{5^{n+1}}+\frac{\cos 5^{n+2}}{5^{n+2}}+\cdots+\frac{\cos 5^{n+p}}{5^{n+p}}\right| \\ & \leq\left|\frac{\cos 5^{n+1}}{5^{n+1}}\right|+\left|\frac{\cos 5^{n+2}}{5^{n+2}}\right|+\cdots+\left|\frac{\cos 5^{n+p}}{5^{n+p}}\right| \\ & \leq\left|\frac{1}{5^{n+1}}\right|+\left|\frac{1}{5^{n+2}}\right|+\cdots+\left|\frac{1}{5^{n+p}}\right|=\frac{1}{5^{n+1}}+\frac{1}{5^{n+2}}+\cdots+\frac{1}{5^{n+p}} \\ &=\frac{1}{5^{n+1}} \frac{1-\left(\frac{1}{5}\right)^{p}}{1-\frac{1}{5}}=\frac{1}{5^{n+1}} \cdot \frac{5}{4} \cdot\left(1-\left(\frac{1}{5}\right)^{p}\right) \end{aligned} $$

Then I'll take the limit of $|U_{n+p}-U_{n}|$ as $n \rightarrow \infty$, which it's zero ; so isn't that enough to prove that the sequence is convergent in $\mathbb{R}$?

$\endgroup$
5
  • 1
    $\begingroup$ That is not sufficient, see math.stackexchange.com/q/1536274/42969 and the linked questions. $\endgroup$
    – Martin R
    Nov 15, 2021 at 10:03
  • 3
    $\begingroup$ The series $\sum \frac {cos (5^{n})} {5^{n}}$ is absolutely convergent by M-test since $|\cos x| \leq 1$. $\endgroup$ Nov 15, 2021 at 10:03
  • 3
    $\begingroup$ Put $a_n=\ln n$. Then $|a_{n+p}-a_n|=|\ln(n+p)-\ln(n)|=|\ln(1+\frac{p}{n})|\to\ln 1=0$ when $n\to\infty$ - yet $(a_n)$ is a divergent sequence. The difference between what you want to use and what you actually need to use for Cauchy criterion is similar to the difference between pointwise and uniform convergence: it is not enough that $|a_{n+p}-a_n|$ becomes arbitrarily small with $n$ big enough, but also that the "moment" when that happens is independent on $p$. $\endgroup$ Nov 15, 2021 at 10:06
  • 1
  • $\begingroup$ thanks folks all makes sense now $\endgroup$ Nov 15, 2021 at 12:40

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.