If $\lim_{n\rightarrow\infty} f(n+1) - f(n) = L$, prove that $\lim_{n\rightarrow\infty} f(n)/n = L$

Solution 1:

To make the indices work nicely, we use a completely unnecessary little trick. Our function $f$ is presumably defined over the positive integers. It is convenient to extend the definition to non-negative integers, by letting $f(0)=0$. Note that $$f(n)=f(n)-f(0)=(f(1)-f(0))+(f(2)-f(1))+\cdots +(f(n)-f(n-1)).$$ Letting $g(n)=f(n)-f(n-1)$, and dividing by $n$, we get $$\frac{f(n)}{n}=\frac{g(1)+g(2)+\cdots +g(n)}{n},\tag{1}$$ We want to show that the right-hand side of (1) has limit $L$. We have $$f(n)-nL=(g(1)-L)+(g(2)-L)+\cdots+(g(n)-L),$$ so by the Triangle Inequality $$|f(n)-nL|\le \sum_1^n |g(i)-L|.\tag{2}$$

Let $\epsilon\gt 0$ be given. There is an $M$ such that if $k\gt M$ then $|g(k)-L|\lt \epsilon/2$. Let $B=\sum_1^M |g(i)-L|$. If $n\gt M$, then the right side of (2) is less than $$B+(n-M)\frac{\epsilon}{2}.$$ Dividing by $n$, we find that for $n$ beyond $M$ we have $$\left|\frac{f(n)}{n}-L\right|\lt \frac{B}{n}+\frac{n-M}{n}\cdot\frac{\epsilon}{2}.\tag{3}$$

Finally, by choosing $n$ large enough and beyond $M$, we can make the right-hand side of (3) less than $\epsilon$. All we need to do is to make $\frac{B}{n}\lt \frac{\epsilon}{2}$.

Remark: We gave a quite detailed proof, as a sample of how to handle the epsilons. The details may hide the basic simplicity of the idea. After we use inverse telescoping to express $f(n)-nL$ as a sum of the $g(i)-L$, the rest is geometrically quite natural.

Solution 2:

Hint : Check the cesaro theorem regarding arithmetical mean of a sequence, when it converges towards a limit L