The question:

Suppose that $f: \mathbb{N} \to \mathbb{R}$. If $$\lim_{n \to \infty}[f(n+1)-f(n)] =L$$ prove that $\lim_{n \to \infty}\frac{f(n)}{n}$ exist and equals $L$.

My answer:

Define $g(n):= f(n+1)-f(n)$ then $g(n-1)=f(n)-f(n-1)$ and so $f(n) = g(n-1)+f(n-1)$. We can recursively apply this to get $f(n)=g(n-1)+g(n-2)+ \cdots +g(0) +f(0)$.

Then for $k \ge N_0$ we can make $|g(k) - L| < \epsilon_0$ which gives $L-\epsilon_0 <g(k)<L+ \epsilon_0$. Also note that $f(N_0) = f(0) + g(0) + \cdots + g(N_0 -1)$ is just some number say $d \in \mathbb{R}$.

Now $f(N_0 + n) = d + g(N_0) + g(N_0 +1) + \cdots + g(N_0+n-1)$ and so using the above bound for $g(k)$ with $k \ge N_0$ we can bound $f(N_0+n)$ as follows:$$d+n(L-\epsilon_0) < f(N_0+n)<d+n(L+\epsilon_0)$$ $$\frac{d}{n} +L-\epsilon_0 < \frac{f(N_0+n)}{n}<\frac{d}{n} +L+\epsilon_0$$ $$\frac{d}{n} -\epsilon_0 < \frac{f(N_0+n)}{n}-L<\frac{d}{n}+\epsilon_0$$

We can take $n$ large enough so that $|\frac{d}{n}| < \frac{\epsilon}{2}$. Also take $\epsilon_0 = \frac{\epsilon}{2}$ and we have shown that the limit of $\frac{f(N_0+n)}{n} = L$.

If we define $h(n) = \frac{f(N_0+n)}{n}$ then $h(x_n)$ will go to $L$ for every $x_n$ that goes to infinity as $n$ does. So take $x_n = n-N_0$ and we have $\lim_{n \to \infty}h(n-N_0) =L= \lim_{n \to \infty}\frac{f(n)}{n-N_0}$ dividing top and bottom by $n$ we see that: $$\lim_{n \to \infty}\frac{\frac{f(n)}{n}}{1-\frac{N_0}{n}}=\lim_{n \to \infty} \frac{f(n)}{n} =L$$

I feel like I have missed something because this question is a lot more involved than all of the others I have done so far in the book. Is this correct? Is there a nicer way to do this? Or have I just wasted two days and rambled nonsense ;-)

  • $\begingroup$ You mean $g(n) = f(n+1) - f(n)$, not $g(x)$. $\endgroup$ – Robert Israel Mar 17 '13 at 18:18
  • $\begingroup$ @RobertIsrael yes i do :-) $\endgroup$ – Ben Mar 17 '13 at 18:20
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    $\begingroup$ It looks fine to me. $\endgroup$ – Brian M. Scott Mar 17 '13 at 18:26
  • $\begingroup$ This is related to the Cesàro mean. $\endgroup$ – Thomas Andrews Mar 17 '13 at 18:36
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    $\begingroup$ (+1) For the effort you've put in and I agree with @Brian's comment. You can turn this in to a $3\epsilon$ proof with the triangle inequality. It is essentially the same approach as you've taken, but might make some things more clear. Just write $$\bigg|\frac{f(n)}{n} - L\bigg| = \bigg|\frac{f(1) - L}{n} + \sum_{k=2}^n \frac{f(k)-f(k-1) - L}{n} \bigg| \leq \frac{|f(1)-L|}{n} + \sum_{k=2}^{N_0} \frac{|f(k)-f(k-1)-L|}{n} + \sum_{k=N_0 + 1}^n \frac{|f(k)-f(k-1)-L|}{n} \>.$$ Now choose $N_0$ appropriately and you are done. $\endgroup$ – cardinal Mar 17 '13 at 18:38

There is this

THEOREM Let $a_n$ be a sequence, and define $\hat{a}_n=\frac 1 n\sum_{k=0}^n a_k$. Then $a_n\to \ell \implies \hat a_n\to\ell$

For a question on that, see here. Use that theorem with $a_n =f(n+1)-f(n)$.

A sequence $a_n$ such that $\lim \hat{a}_n$ exists is called Cesàro summable. Thus, the above claims all convergent sequences are Cesàro summable. The converse is not true. For example, Grandi's series $1,0,1,0,\dots$ does not converge, however, it is Cesàro summable with sum $1/2$.

  • $\begingroup$ Is it different from Sami's answer? It is Cesàro theorem. $\endgroup$ – Mhenni Benghorbal Mar 17 '13 at 19:27

By Cesàro theorem we have $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n-1}f(k+1)-f(k)=\lim_{n\to\infty}\frac{f(n)-f(1)}{n}=\lim_{n\to\infty}\frac{f(n)}{n}=L.$$

  • $\begingroup$ Yes, right. Thanks. $\endgroup$ – user63181 Mar 17 '13 at 19:15

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