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

Question

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.
Thank you in advance for your answers.

Answer 1

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.

Answer 2

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 $0=n_0<n_1<n_2<\cdots$.

Now, define $b_n=(-1)^k a_n$ for $n_{k-1}<n\leq n_k$. 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 $v_n=\sum_{k<n} b_k$ 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.