# $\text { Prove } u_{n}=\frac{\cos 5}{5}+\frac{\cos 5^{2}}{5^{2}}+\cdots+\frac{\cos 5^{n}}{5^{n}} \text { is convergent in } \mathbb{R} \text {. }$

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 $$\mathbb{R}$$ ;

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 $$\mathbb{R}$$?

• That is not sufficient, see math.stackexchange.com/q/1536274/42969 and the linked questions. Nov 15, 2021 at 10:03
• The series $\sum \frac {cos (5^{n})} {5^{n}}$ is absolutely convergent by M-test since $|\cos x| \leq 1$. Nov 15, 2021 at 10:03
• 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 $p$. Nov 15, 2021 at 10:06
• Nov 15, 2021 at 10:08
• thanks folks all makes sense now Nov 15, 2021 at 12:40