Show that $f(x)=\sum_{n=1}^\infty 1/n^x$ is not uniformly continuous on $(1,\infty)$?

Question

Show that $f(x)=\sum_{n=1}^\infty 1/n^x$ is not uniformly continuous on $(1,\infty)$? How can I do that? May I argue by contradiction. Then what?

Answer 1

Use the inequalities:

$$\int_1^\infty \frac{1}{n^x} dn < \sum_{n=1}^\infty \frac{1}{n^x} < \int_1^\infty \frac{1}{n^x} dn + 1$$

To see this, note that $\displaystyle \frac{1}{n^x}$ is monotonic decreasing for $x > 1$, and think of this image:

Now note that $$\int_1^\infty \frac{1}{n^x} dn = \frac{1}{x-1}$$

Now exploit the fact that $\displaystyle \frac{1}{x-1}$ is not uniformly continuous on $( 1 , \infty )$ (look at values very close to $1$). Mouse over below to see a fuller proof:

Take $\epsilon = 1$. Then for every $\delta$ (we can assume, without loss of generality, that $\delta < 1$), take $x = 1+\delta / 2$ and $y = 1 + \delta$. This gives us $d(f(x),f(y)) \ge \frac{2}{\delta}-1 > 2-1 = \epsilon$.