1
$\begingroup$

$f_{k}(x) =\frac{1}{ln(1+kx)} , k\geq 1$

a)x>1 b)x>0

My attempt : a) because for x>0 the natural logaritm is a monotonic decreasing function the sequence converges to 0.

b) $M_{k} = Sup_{x>0} \frac{1}{ln(1+kx)}$ , for $x =\frac{1}{k}$ $M_{k}=\frac{1}{ln2}\not\Rightarrow 0$ Since M_k does not go zero it does not converge uniformly.

Can you please tell me if am correct or not ? am new on this topic , a detailed answer would be very helpfull

Thanks in advance

  • 0
    I'm not sure what you did in $a)$ but your reasoning in $b)$ is correct. For $a)$, note that for $x>1$, $f_k(x) \le \frac{1}{\ln(1+k)}$, so $\sup_{x>1} f_k(x) \le \frac1{\ln(1+k)} \to$, showing uniform convergence to $0$. – 2017-01-05

0 Answers 0