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Suppose that f is a bounded function from R to R. Prove that $\frac{f(n)}{n}$ $n=1 \to \infty$ converges.

Normally, I would just fix epsilon > 0 then try to find rearrange the inequality of the absolute value of the given series minus its limit < epsilon to find an n to satisfy the problem. However, I can't exactly do that here seeing how f(n) could be any function. How do I do this?

Attempt after hint:

|f(n)/n - 0| < epsilon

= |f(n)/n| < epsilon

=|f(n)|/n < epsilon

= M/n < epsilon

= n < M/epsilon

Fix epsilon > 0. Let N = ceiling function (M/epsilon) where M is some constant. Then if n >= N

We have |f(n)/n - 0| = |f(n)/n| <= 1/n <= 1/N < epsilon

I'm actually quite confused on this problem.

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    First note that a bounded sequence in R has at least on accumulation point. Then you still have to show that there is only one accumulation point (with some epsilon argument) – 2017-02-16

1 Answers 1

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Hint:

$|f(n)|\leq M$ for some constant $M$.

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    was my edit what you meant when you gave me the hint? – 2017-02-16