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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.

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    $\begingroup$ Let $u_n>0$ and take log changes the problem to something that should look familiar. $\endgroup$ Oct 20, 2018 at 10:16
  • $\begingroup$ 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. $\endgroup$
    – Tengen
    Oct 20, 2018 at 10:19
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    $\begingroup$ 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. $\endgroup$ Oct 20, 2018 at 10:29

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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.

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  • $\begingroup$ Thank you for your answer ! $u_{n+1}-u_n \rightarrow 0$ as a consequence of the mean value inequality so we just take $e^{2+sin\sqrt{n}}$ $\endgroup$
    – Tengen
    Oct 20, 2018 at 10:27
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    $\begingroup$ 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 $u_n>1$) $\endgroup$ Oct 20, 2018 at 10:40
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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.

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  • $\begingroup$ 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 $|a_n|<1$. $\endgroup$
    – James Well
    Nov 18, 2019 at 8:57

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