$k_n(x) = \frac x{1+nx^2}$ uniform convergence

Question

Prove that the sequence $\{k_n\}_{n=1}^\infty$ defined by $$k_n(x) = \frac x{1+nx^2}$$ for all $x \in R$ and each positive integer $n$, converges uniformly on R. $$$$ I know the definition of uniform convergence, $\forall\epsilon>0, \exists N \in J$, such that $\forall n \ge N, \forall x \in D$ implies $|k_n(x) - k(x)| < \epsilon$.

I'm just not sure how to find $k(x)$ which this sequence converges to in order to complete the necessary inequality.

Any help would be greatly appreciated! Thanks!

You might want to see this, posted one hour ago http://math.stackexchange.com/questions/1045996/does-the-sequence-f-nx-dfracx1nx2-converge-uniformly-on-mathbbr — Matthew Cassell, Dec 01 '14 at 05:56

score 2 · Accepted Answer · answered Dec 01 '14 at 05:57

2

$k_n(x)=\frac{x}{1+nx^2}<\frac{x}{nx^2}=\frac{1}{nx} \rightarrow 0 $ as $n\rightarrow \infty $ for a fixed $x\in \mathbb R$

So $k(x)=0$

answered Dec 01 '14 at 05:57

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31,062

Yeah that's what I thought, but I wasn't sure. Thank you! – jlang Dec 01 '14 at 06:00
Can you let $N = \frac x\epsilon$? – jlang Dec 01 '14 at 06:03
Should separate the case when $x = 0$, but it is well-understood what you meant. – IAmNoOne Dec 01 '14 at 06:27

score 0 · Answer 2 · answered Aug 14 '17 at 15:14

0

See $M_{n}=\sup \vert k_{n}(x)-k(x) \vert=\frac{1}{\sqrt n}$ for all $x\in \mathbb{R}$ and $n$ tending to $\infty, M_{n}$ goes to zero .

answered Aug 14 '17 at 15:14

jinx

85

$k_n(x) = \frac x{1+nx^2}$ uniform convergence

2 Answers2