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Let $f:(0,1] \rightarrow \mathbb{R}$ be differentiable on $(0,1]$, with $|f'(x)| \leq 1$ for all $x$ in $(0,1].$ For each $n$ in $\mathbb{N}$, let $a_n=f(1/n)$ Show that $(a_n)_{n \in \mathbb{N}}$ converges.

This is what I have so far:

Since $f$ has a bounded derivative on $(0,1]$, $f$ is uniformly continuous on $(0,1]$.

so the definition to be uniformly continuous is $\forall \epsilon >0, \exists \delta >0$ such that if $x,y \in (0,1]$ and $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$.

Now I have to apply the mean value theorem to show $a_n$ is cauchy, thus convergers.

This is where i am stuck. I know the definition of a cauchy sequence is $\forall \epsilon >0,$ $\exists n_0 \in \mathbb{N}$ such that if $n,m \geq n_0$ then $|x_n-x_m|<\epsilon$. How do i put this all together. Thanks for the help!

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1 Answer 1

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The MVT says: $|a_m - a_n| = |f'(c)|\left|\dfrac{1}{m} - \dfrac{1}{n}\right|\le \left|\dfrac{1}{m}-\dfrac{1}{n}\right|< \epsilon$ for $m,n > N_0$ as this $N_0$ exists since $\{\dfrac{1}{n}\}$ is convergent hence is Cauchy. Thus $\{a_n\}$ is a Cauchy sequence hence convergent.

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  • $\begingroup$ oh wow thats what I wrote on my paper but i couldnt think of why $|1/m-1/n|<\epsilon$ is true. But its because both $1/m$ and $1/n$ are cauchy and convergent $\endgroup$
    – jimm bo
    Commented Nov 13, 2016 at 6:38

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