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Been struggling with this for a while now, just kind of spinning my wheels at this point. Any help/advice/anything much appreciated. This is a problem from Wade's Intro to Analysis, chapter 3.

Suppose $f : \textbf{N} \to \textbf{R}$. If $\lim\limits_{n \to \infty}f(n + 1) - f(n) = L$, prove that $\lim\limits_{n \to \infty}\frac{f(n)}{n}$ exists and equals $L$.

daniel
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    The most classical epsilon-delta proof works (or rather, in this case, epsilon-N). – Did Mar 07 '13 at 04:25
  • stolz theorem http://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem – noname1014 Mar 07 '13 at 04:27
  • Did: yes I am attempting to use epsilon-N, I understand that it can be done that way, it is the 'how' that I am stuck on. I have been trying to construct a string of inequalities that leads from the assumption to what I'm trying to prove, but I'm just not getting anywhere material. – daniel Mar 07 '13 at 05:04

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Suppose that we are given an $\epsilon \gt 0$. Then there is a $b=b(\epsilon)$ such that if $k\ge b$, then $f(k+1)-f(k)$ is within $\epsilon/3$ of $L$. This $b$ will be fixed from now on.

Note that $$(f(b+1-f(b))+(f(b+2)-f(b+1))+\cdots +(f(n)-f(n-1))=f(n)-f(b).$$ On the left side we have $n-b$ terms whose sizes are under control Thus $$(n-b)(L-\epsilon/3) \lt f(n)-f(b) \lt (n-b)(L+\epsilon/3).$$ Add $f(b)$ to each of the three sides, and divide by $n$. We get $$\frac{n-b}{n}(L-\epsilon/3)+\frac{f(b)}{n}\lt \frac{f(n)}{n} \lt \frac{n-b}{n}(L+\epsilon/3)+\frac{f(b)}{n}.$$ We conclude that for large enough $n$, $\frac{f(n)}{n}$ is within $\epsilon$ of $L$. It suffices to make sure that $\frac{f(b)}{n}$ is within $\epsilon/3$ of $0$, and that the term $\frac{n-b}{n}$ is close enough to $1$.

André Nicolas
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Suppose $a_n \to L$. Let $s_n = \frac{1}{n}(a_1+\cdots+a_n)$, then we have $s_n \to L$.

Now let $a_n = f(n+1)-f(n)$. Then $s_n = \frac{1}{n}(f(n+1)-f(1))$, and since $\frac{f(1)}{n} \to 0$, we have $\frac{f(n+1)}{n} \to L$, and since $\frac{n}{n+1} \to 1$, we have $\frac{f(n)}{n} \to L$.

Aside: To see that $s_n \to L$, let $\epsilon>0$, and choose $N$ such that $|a_n-L| < \frac{\epsilon}{2}$. Let $n \ge N$ and consider $|s_n-L| \le \frac{1}{n}\sum_{k=1}^n |a_k-L| = \frac{1}{n}\sum_{k=N}^n |a_k-L| + \frac{1}{n}\sum_{k=1}^N |a_k-L| < \frac{\epsilon}{2} + \frac{1}{n}\sum_{k=1}^N |a_k-L|$. Now choose $N' \ge N$ such that if $n \ge N'$, then $\frac{1}{n}\sum_{k=1}^N |a_k-L| < \frac{\epsilon}{2}$. Then if $n \ge N'$, we have $|s_n-L| < \epsilon$.

copper.hat
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    Wow, great response, I've poured over this and have really learned a lot. I have only one question. Why does knowing that $\frac{n}{n+1} \to 1$ lead to the conclusion? – daniel Mar 07 '13 at 07:32
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    If $b_n \to b$ and $c_n \to c$, then $b_n c_n \to bc $. So, since $\frac{f(n+1)}{n} \to L$ and $\frac{n}{n+1} \to 1$, then $\frac{f(n+1)}{n}\frac{n}{n+1} = \frac{f(n+1)}{n+1} \to L\cdot 1 = L$. (And $\frac{f(n+1)}{n+1}$ and $\frac{f(n)}{n}$ have the same limit, of course.) – copper.hat Mar 07 '13 at 07:38
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    dang, that is smooth (that whole proof). – daniel Mar 07 '13 at 08:12
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    It follows years of non-smoothness :-). – copper.hat Mar 07 '13 at 08:17
  • @copper.hat: great. Could you (if possible) explain a bit more the last part: $\sum_{k=1}^{N}|a_k-L|<\frac{\epsilon}{2}$? – Alex Nov 13 '13 at 01:45
  • @Alex: It is $\frac{1}{n} \left( \sum_{k=1}^{N}|a_k-L| \right) < \frac{\epsilon}{2}$. It follows because $a_k \to L$, so the $a_k$ are bounded. Then pick $n$ large enough. – copper.hat Nov 13 '13 at 01:55
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Take An=f(n+1)-f(n) then lim An= L implies lim (A1+A2+....+An)/n=L (BY Cauchy's first principle) i.e. lim {f(n+1)-f(1)}/n=L i.e. lim {f(n+1)}/n=L

Also lim {f(n+1)-f(n)}/n=lim L/n = 0 which gives lim {f(n+1)}/n=lim{f(n)}/n Hence the solution.

mukesh
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