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Looking to show that the following sequence $(f_n)$ converges uniformly and then finding its limit.

$$f_n(x)=\frac{x^n}{n}$$ for $x\in[0,1]$

Converges uniformly:

My thinking here was to use the M test to prove it converges uniformly,

since $x\in[0,1]$ if we choose $x<=1$ and let R be 1, then,

$$\frac{x^n}{n}<\frac{R^n}{n}$$ but now I would have to show that $\frac{R^n}{n}$ converges?

If this logic is correct how do I proceed with proving that it converges uniformly? and then how would I find its limit?

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    Check the different cases $x = 0$, $x = 1$, $x \in (0, 1)$ individually. What happens in the limit in each case? – 2017-02-28
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    @Mattos when x=0 then the limit is 0, when x=1 then the limit is 0, but what about (0,1?? – 2017-02-28
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    For $x \in (0, 1)$, $x^{n}$ is a decreasing sequence as $n \to \infty$. – 2017-02-28
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    @Mattos so would the limit be 0 then? – 2017-02-28

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Remember that $\frac{R^n}{n} = \frac{1}{n}$. As $n$ goes to $\infty$, what does this converge to?

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    converges to 0? but isn't 1/n a divergent p series? – 2017-02-28
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    @user123 It is a divergent series but a convergent sequence. The difference is whether you are adding up all of the terms. – 2017-02-28