test

**becko** · April 24th 2010, 10:26 AM

$<br /> {a}_{n}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}<br /> =\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{n} }\frac{1}{k}<br /> <\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{{k}^{3/2}}<br />$

**becko** · April 25th 2010, 09:14 AM

Let $x,t\in \left(a,b \right)$ .

By the mean value theorem of derivatives:

$f(x)-f(t)=(x-t)f'(\lambda )$

where $\lambda \in (x,t)$ . Hence:

$\left| f(x)-f(t)\right|\leq M\left| x-t\right|$

Now for any $\epsilon >0$ , we can put

$\delta =\frac{\epsilon }{M}$

so that whenever $\left| x-t\right|<\delta$ , we will have $\left| f(x)-f(t)\right|<\epsilon$ . This proves that $f(x)$ is uniformly continuous.

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