While solving some exercises for my students, I bumped into the following sub-problem: given a divergent sequence $$(a_n)$$, such that $$(a_{n+1} – a_n)$$ converges to $$1$$, can we state that $$a_n \sim n$$ for $$n \to +\infty$$? The answer to the quesiton, as I propose it, requires some drawing, and it is useful to practice the definition of limit.

Theorem. Let $$(a_n$$) be a diverging sequence such that$\lim_{n\to +\infty} (a_{n+1}-a_n) = 1,\tag{1}\label{eq3531:1}$ then we have$\lim_{n\to +\infty} \frac{a_n}{n} = 1.$

For any given $$\varepsilon > 0$$, we need to find $$N_\varepsilon>0$$ such that, for all $$n> N_\varepsilon$$, $\left|\frac{a_n}{n}-1\right|<\varepsilon\tag{2}\label{eq3531:2}.$

Note that \eqref{eq3531:2} is equivalent to stating that, for large enough $$n$$, the points $$P_n(n,a_n)$$ must lie between the straight lines $$r_1$$ and $$r_2$$ with equations $$y=(1+\varepsilon) x$$ and $$y=(1-\varepsilon) x$$, respectively. Let us call this region $$\mathcal Q_\varepsilon$$.

By \eqref{eq3531:1}, we can state that there exists $$N$$ such that, for $$k\geq N$$, $a_k+1-\frac{\varepsilon}2 < a_{k+1}< a_k+1+\frac{\varepsilon}2.\tag{3}\label{eq3531:3}$

Repeated application of \eqref{eq3531:3} for $$k=N,N+1,\dots,N+n$$, yields$a_N + n – n\frac{\varepsilon}2< a_{N+n} < a_N + n + n\frac{\varepsilon}2,\tag{4}\label{eq3531:4}$ valid for all $$n>0$$. Again it is useful to interpret \eqref{eq3531:4} geometrically. The above equation states that for $$n>0$$ the points $$P_{N+n}(N+n,a_{N+n})$$ are contained within the straight lines with equations $s_1 : \ \ y=(x-N)\left(1+\frac{\varepsilon}2\right) + a_N$ and$s_2 : \ \ y=(x-N)\left(1-\frac{\varepsilon}2\right) + a_N$(this corresponds to the shaded area in the Figures below).

The slopes of these lines have been chosen in such a way that, if $$P_N \in \mathcal Q_\varepsilon$$, so are all points $$P_n$$ for $$n>N$$. This allows us to simply take $$N_\varepsilon = N$$. See Figure below. Note that in this case the intersections of lines $$r_1$$ and $$r_2$$ with $$s_1$$ and $$s_2$$ all occur at $$x < N$$.

If $$P_N$$ lies above $$r_1$$, then we must find the intersection between $$r_1$$ and $$s_1$$. This leads to $N_\varepsilon = \left\lceil \frac{2}{\varepsilon}(a_N – N)-N\right\rceil.$This situation is depicted in the following Figure.

Similarly, if $$P_N$$ lies below $$r_2$$ we need to find the intersection between $$r_2$$ and $$s_2$$, leading to $N_\varepsilon = \left\lceil \frac{2}{\varepsilon}(N -a_N)-N\right\rceil.$

In conclusion, once $$N$$ has been chosen according to condition \eqref{eq3531:3}, selecting $N_\varepsilon = \max\left\{N, \left\lceil \frac{2}{\varepsilon}(a_N – N)-N\right\rceil, \left\lceil \frac{2}{\varepsilon}(N – a_N)-N\right\rceil\right\}$guarantees \eqref{eq3531:2}, and the thesis follows.

(Putnam 2006, Exercise B.6). Let $$k$$ be an integer greater than $$1$$. Suppose $$a_0 > 0$$, and define $a_{n+1} = a_n + \frac{1}{\sqrt[k]{a_n}}$for $$n > 0$$. Evaluate $\lim_{n \to \infty} \frac{a_n^{k+1}}{n^k}.$

1. The sequence $$(a_n)$$ is monotonically increasing, therefore it must have a limit, either finite or infinite. If $$\lim_{n\to\infty} a_n = a<\infty$$, then $$(a_{n+1}-a_n)$$ is a null sequence. Show that this generates a contradiction. Thus the sequence diverges.
2. Let $$b_n = a_n^{\frac{k+1}k}$$. (Note that $$b_n$$, too, is divergent.) Using the fundamental limit $\frac{(1+\alpha)^m-1}{\alpha} \to m$ when $$\alpha \to 0$$, show that $b_n-b_{n-1} \to \frac{k+1}k$ when $$n\to \infty$$.
3. Use this result, and the Theorem we just proved, to show that, for $$n\to \infty$$, $\frac{b_n}{n\cdot\frac{k+1}k} \to 1.$
4. Conclude that $\lim_{n \to \infty} \frac{a_n^{k+1}}{n^k} = \left(\frac{k+1}k\right)^k.$