# Pointwise convergence of $\frac{n}{xn+1}$

For each $$n ∈ \Bbb N$$, let $$f_n : (0, 1) \to \Bbb R$$ be defined as`$$f_n(x) = \frac{n}{xn+1}$$ Prove $$(f_n )^{∞}_{n=1}$$ converges pointwise on $$(0,1)$$

That is, the $$f_n$$'s go to zero

I tried finding the limit as n approaches infinity of $$\frac{n}{nx+1}$$, but that is equal to $$\frac{1}{x}$$, and given that $$x ∈ (0,1)$$, this seems contrary to the idea that the sequence converges pointwise. What am I overlooking?

• You have a sequence, not a series. – egreg Dec 5 '18 at 11:04

$$f_n$$'s do not go to zero. What you do is correct. We fix $$x\in(0,1)$$ and then we investigate the limit as $$n\to\infty$$. We have that, for each fixed $$x\in(0,1)$$, $$f_n(x)\to1/x$$ as $$n\to\infty$$. Hence, $$f_n$$ converges pointwise to $$f$$ as $$n\to\infty$$, where $$f$$ is defined by $$f(x)=1/x$$ on $$(0,1)$$.