# $\frac{u_{n+1}}{u_n}\rightarrow 1$ but $(u_n)$ does not have a limit in $\bar{\mathbb{R}}$

I've been struggiling to find a real sequence $$(u_n)$$ such that $$\frac{u_{n+1}}{u_n}\rightarrow 1$$ and $$(u_n)$$ does not have a limit in $$\bar{\mathbb{R}}=\mathbb{R}\cup \left\{-\infty ,+\infty \right\}$$. I can't find one.

• Let $u_n>0$ and take log changes the problem to something that should look familiar. Oct 20, 2018 at 10:16
• That changes the question to finding a sequence $(v_n)$ such that $v_{n+1}-v_{n}\rightarrow 0$ and $(v_n)$ has no limit in $\bar{\mathbb{R}}$, but I still can't find an example of such a sequence. Oct 20, 2018 at 10:19
• Letting $a_n=v_{n+1}-v_n$ gives the equivalent formulation $a_n\to 0$ but the sequence of partial sums $\sum_{j=1}^n a_j$ does not converge in $\bar{\mathbb{R}}$, which you should have seen before. Oct 20, 2018 at 10:29

Try something like $$\tag1u_n=2+\sin\sqrt n.$$ Note that $$u_n\ge 2-1=1$$ for all $$n$$ and that $$\tag2\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\to 0$$ so that \begin{align}u_{n+1}-u_n&=\sin\sqrt{n+1}-\sin\sqrt n \\&=2\cdot \underbrace{\cos\frac{\sqrt{n+1}+\sqrt n}2}_{|\cdot|\le 1}\cdot \sin \underbrace{\frac{\sqrt{n+1}-\sqrt n}2}_{\to 0}\\&\to 0\end{align} and therefore $$\frac{u_{n+1}}{u_n}=1+\frac{u_{n+1}-u_n}{u_n}\to 1,$$ whereas $$\sqrt n\to\infty$$ together with $$(2)$$ implies $$u_n\approx 3$$ as well as $$u_n\approx 1$$ infinitely often.

• Thank you for your answer ! $u_{n+1}-u_n \rightarrow 0$ as a consequence of the mean value inequality so we just take $e^{2+sin\sqrt{n}}$ Oct 20, 2018 at 10:27
• Actually, I suggested to take $u_n=2+\sin\sqrt n$ without modification. Note that $\frac{u_{n+1}}{u_n}=1+\frac{u_{n+1}-u_n}{u_n}\to 1$ as the numerator $\to0$ and the denominator is bounded away from $0$ (as $u_n>1$) Oct 20, 2018 at 10:40

Take any sequence $$a_n$$ of positive real numbers such that

1. $$\sum_n a_n$$ diverges to infinity.
2. $$a_n$$ converges to 0.

(for example $$a_n=1/n$$).

Next, define $$n_k$$ (starting with $$n_0=0$$) so that $$\sum_{n>n_{k-1}}^{n_k} a_n$$ lies in the interval $$(k,k+1]$$. It is clear that $$0=n_0.

Now, define $$b_n=(-1)^k a_n$$ for $$n_{k-1}. The series $$\sum_k b_k$$ does not converge since the partial sums "go left" until they cross -1, then "go left" until they cross -1 and so on. Each time, the partial sums change direction, they cross $$\pm 1$$ before they change direction again.

Now, the partial sums $$v_n=\sum_{k are such that:

1. $$v_{n+1}-v_n$$ converge to 0.
2. $$v_n$$ do not converge.

Taking $$u_n=\exp{v_n}$$ will do what you need.

• I think you may not be able to define $n_k$ starting at $n_0$ if $(a_n)_n$'s first terms are too large, but only beyond the point where every $|a_n|<1$. Nov 18, 2019 at 8:57