A seemingly true claim?

Question

Let $(a_n)$ be a sequence of positive real numbers. If $(a_n)$ is not bounded then $\lim\sup\Big(\frac{a_n}{1+a_n}\Big)=1$.

I was tempted to make the above claim when I tried proving the result below:

Let $(a_n)$ be a sequence of positive real numbers. If $(a_n)$ is not bounded then show that $\Big(\frac{a_n}{1+a_n}\Big)$ does not tend to zero.

My attempt :

If $(a_n)$ is not bounded then there exists a strictly increasing(?) subsequence $(a_{n_k})$ of $(a_n)$ such that $a_{n_k}\to \infty.$

Claim : $\Big(\frac{a_{n_k}}{1+a_{n_k}}\Big)$ is a positive monotonically increasing sequence.

\begin{align} \frac{a_{n_{k+1}}}{1+a_{n_k+1}}-\frac{a_{n_k}}{1+a_{n_k}}=\frac{a_{n_{k+1}}-a_{n_k}}{(1+a_{n_k+1})(1+a_{n_k+1})}> 0 \hspace{3mm}(\because a_{n_{k+1}}-a_{n_k}>0) \end{align}

Hence $\Big(\frac{a_{n_k}}{1+a_{n_k}}\Big)$ is monotonic. Also it is evident that this is a positive sequence.

Since there exists a subsequence which does not tend to zero implies that $\Big(\frac{a_n}{1+a_n}\Big)$ does not tend to zero.$\hspace{15cm} \blacksquare$

Looks to me that this particular subsequence has no reason to converge to any number other than $1.$ Is my claim true?

Answer 1

$\lim_{x \to \infty} \frac x{1+x} =1$. So if $a_{n_k} \to \infty$ then $\frac {a_{n_k}} {1+a_{n_k}} \to 1$.

Since $\frac {a_n} {1+a_n} <1$ for all $n$ we can conclude that the limit superior is $1$.

Answer 2

Just sharing my opinion.Assume the contrary that, $x_n=\frac{a_n}{1+a_n}$ tends to zero, then for any fixed $0<\epsilon<1$ there is a $N\in\Bbb N$ such that, $x_n<\epsilon,\forall n\ge N$, which gives $\displaystyle a_n<\frac{\epsilon}{1-\epsilon},\forall n\ge N$. Now take $M=max\{\frac{\epsilon}{1-\epsilon},a_1,...,a_{N-1}\}$, we see $a_n<M,\forall n\in\Bbb N$ i.e. Bounded-a contradiction that $\{a_n\}$ is not bounded. So the sequence $\{x_n\}$ doesnot tends to zero.