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I'm posting this question just to be sure if my solution is correct or not.

I have a sequence $\{f_{n}\}$, where each $f_{n}:\mathbb{R}\to\mathbb{R}$ is continuous on $\mathbb{R}$ and $|f_{n}(x)|< a_{n}$ for all $n=1,2,3,\dots$ , for some sequence $a_{n}$ converges to 0. I'm trying to prove that $\frac{|f_{n}(x)|}{a_{n}}\to 0$ as $n\to\infty$.

I started with: since $|f_{n}(x)|\to 0$ as $n\to\infty$ (because $a_{n}\to 0$) for all $x\in\mathbb{R}$, this means that:

for any $\epsilon>0$ there exists $N_{\epsilon}>0$ such that $|f_{n}(x)|< \epsilon$ for all $n> N_{\epsilon}$ and all $x\in \mathbb{R}$, (uniform convergence)

Also, we can say: for any $j\in \mathbb{N}$ (by taking $\epsilon=a_{j}^{2}$) there exists $N_{j}>0$ such that $|f_{n}(x)|< a_{j}^{2}$ for all $n> N_{j}$ and all $x\in \mathbb{R}$, therefore: $$\frac{|f_{n}(x)|}{a_{n}}<\frac{a_{j}^{2}}{a_{n}} \to 0,\; as \;j\to \infty$$ for all $n>N_{j}$ and all $x$.

So, for large $n$ we have $\frac{|f_{n}(x)|}{a_{n}}\to 0$ by taking the limit as $j\to \infty$.

I think there is something wrong somewhere in my argument, if so please correct me!

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    Just a note: lets forget about constant functions like $f_{n}(x)=1/(n+1)$ (since we will get $1/(n+1)< 1/n$ and the limit will be 1 not 0).2012-06-19
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    The argument that went wrong in your proof is that when letting $j \rightarrow \infty $ then in the inequality your $n$ is $n(j)$ it depends on $j$ so as far as we know $n$ could be must larger than $j$ thus much nearer to $0$ and that means $ \frac{a_{j}^2}{a_n}$could be big2012-06-19
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    We usually can get one of the indices to tend to $\infty$ when they are independent. For example if you have something like that $\forall m,n \geq n_0$ we have some inequality then we can make $m$ or $n$ tend to $\infty$2012-06-19
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    @Dimon: You should accept the answer below, as it completely answers your question.2012-06-21

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