test

• April 24th 2010, 11:26 AM
becko
test
$
{a}_{n}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}
=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{n} }\frac{1}{k}
<\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{{k}^{3/2}}
$
• April 25th 2010, 10:14 AM
becko
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.