Does $f_k(x):=\frac{1}{x^{1+\frac{1}{k}}}$ converge uniformly?

Question

Let be $f_k:[1,\infty)\to\mathbb{R}$ with $f_k(x):=\frac{1}{x^{1+\frac{1}{k}}}$. Does this sequence converge uniformly?

My approach:

It is easy to see that $(f_k)_{k\in\mathbb{N}}$ converges pointiwsely to $f:=\frac{1}{x}$. Also $$ \left|\frac{1}{x^{1+\frac{1}{k}}}-\frac{1}{x}\right|\leq \frac{\left|x^{\frac{1}{k}}-1\right|}{x}. $$ Though it seems that the upper bound goes to $0$ independent from $x$, which would allow me to say something like $\sup\{|f_k(x)-f(x)|\mid x\in[1,\infty)\}=0$, I am not able to carry it out rigorously.

Do you have any hints? Maybe there is a tricky manipulation which might help?

Answer 1

For $x \ge 1$ and $k \ge 1$ is, using Bernoulli's inequality, $$ x \le x^{1+1/k} = x (1+(x-1))^{1/k} \le x \left(1 + \frac{x-1}{k}\right) \, . $$ It follows that $$ 0 \le \frac 1x - \frac{1}{x^{1+1/k}} \le \frac1x \left( 1 - \frac{1}{1+(x-1)/k}\right) = \frac{x-1}{x(x-1+k)} \le \frac 1k $$ and that proves the uniform convergence on $[1, \infty)$.

We can also start with your estimate: $$ \left|\frac{1}{x^{1+\frac{1}{k}}}-\frac{1}{x}\right|\leq \frac{x^{\frac{1}{k}}-1}{x} = \frac{(1+(x-1))^{1/k}-1}{x} \le \frac{(x-1)/k}{x} \le \frac 1k \, $$ again using Bernoulli's inequality.

Answer 2

Your bounding functions, $g_k(x):= \frac{x^{1/k}-1}{x}$ is differentiable in the domain and attains the maximum in some interior point $x_k$. Differentiating it gives you that $x_k= \big( \frac{k}{k-1} \big)^{k}$.

Therefore, $\sup_{x\geq1} g_k(x)=g_k(x_k)$. You just have to check whether $g_k(x_k)\to0$. You can see that

$$ g_k(x_k)= \frac{ \big( \frac{k}{k-1} \big)^k-1 }{\big( \frac{k}{k-1} \big)^k}= \frac{ \frac{1}{k-1} }{ \big( 1+ \frac{1}{k-1} \big)^{k-1} \cdot \big( \frac{k}{k-1} \big)}\to \frac{0}{e\cdot 1}=0. $$

This gives you another way to show uniform convergence.