Why is the series $\sum_n\frac{1}{n^x}$ not uniformly convergent on $x\in(1, \infty)$ I've been struggling with this problem for the past 5 days. I've tried to prove the series $\sum_{n=1}^\infty\frac{1}{n^x}$ is not uniformly convergent when $x$ belongs to $(1,\infty)$ but to no avail. Using Cauchy's criterion where $n>m$, if I can prove that it is not uniformly Cauchy then I will have proved it is not uniformly convergent. Then, $|f_n(x)-f_m(x)|= (\frac{1}{m+1})^x + ... + \frac{1}{n^x} > \frac{n-m}{n^x} \geq \frac{1}{n^x} > L$ when $x<\frac{\ln(\frac{1}{L})}{\ln(n)}$ but that would hold for any $n$ no matter how big only if we are to consider $x<1$ but here the interval is $(1,\infty)$. Any hint on how to prove this? 
I've also managed to prove using Cauchy's criterion that the same sequence is Cauchy for any interval $(x,\infty)$ where $x>1$, which sounds contradictory.
Please help!!
 A: Hint: A series with bounded terms can't converge uniformly to an unbounded result.
A: If the series $\displaystyle \sum_{n=1}^\infty n^{-x}$ were to converge uniformly to its limit on the interval $(1,\infty)$, it would have to be uniformly Cauchy. In particular there would be an index $N$ with the property that 
$$ k > m \ge N \implies \sum_{n=m}^k n^{-x} \le 1,\quad x \in (1,\infty).$$ 
Since the function on the left hand side of the inequality is continuous for $x > 0$ it follows by taking $x \to 1^+$ that $$k > m \ge N \implies \sum_{n=m}^k n^{-1} \le 1.$$
However, we may select $K > N$ with the property that 
$$\sum_{n=N}^K n^{-1} > 2$$ because the harmonic series diverges. This contradiction indicates that the series does not converge uniformly.
A: The sum $\sum\frac1n$ diverges to infinity, therefore for any $m$ the sum $\sum_{k=m}^n\frac1{k}$ can be made arbitrarily large with large enough $n$.  And if $\sum_{k=m}^n\frac1{k}>M$ then since $\sum_{k=m}^n\frac1{k^x}$ is a continuous function of $x$ on $[1,\infty]$ (for fixed $m$ and $n$), you can find an interval $[1,1+\delta]$ so that $\sum_{k=m}^n\frac1{k^x}>M/2$ for all $x\in[1,1+\delta]$.  It should then follow that $\sum_{k=1}^n\frac1{k^x}>M/2$ for some $x>1$ and so the series cannot be uniformly Cauchy  on $(1,\infty)$.
