How do I find the upper bound of the sequence $\left(a_n\right)_n$ with $a_n=\frac{2n}{1+n}$ I recently started to study sequences and series in calculus and studied the concept of bounded sequences. In the following seuquence: $\left(a_n\right)_n$ with $a_n=\frac{2n}{1+n}$
I can see clearly that the lower bound is 1. But I am perplexed when it comes to finding the upper bound.
I can't plug in infinity because I would get infinity over infinity. So I try to apply L'Hopital's rule and I find that the derivative of the numerator over derivative of the denominator is 2.
Is this enough to conclude that the upper bound is 2?
 A: You need a bit more. That the limit of the sequence is $2$ isn't enough. You also need, that this sequence is monotonous. Then you can say that $2$ is the upper bound.
You can try it this way:
As $n$ approaches infinity, you see that $n$ gets much bigger than 1, so you can neglect it and you get
$$ \lim_{n\rightarrow\infty} \frac{2n}{1+n} = \lim_{n\rightarrow\infty}\frac{2n}{n} = 2. $$
Because $a_n$ is monotonous, you see, that this is indeed the upper bound.
A: Firstly, it it easily shown that the sequence is monotonically increasing. As a consequence, if it converges, It's upper bound is identified with the limit.
Given that you calculate:
$$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{\frac{1}{n}(2n)}{\frac{1}{n}(1+n)}=\lim_{n \to \infty}\frac{2}{\frac{1}{n}+1}=\frac{2}{0+1}=2$$
A: Adding to @Alex R:
$$\lim_{n\to\infty} \frac{2n}{1+n} \stackrel {\text{*}}= \lim_{n\to\infty}\frac{2}{\frac1n+1} = 2.$$
*Divide and multiply by $n$
Also, in general, L'Hôpital is kind of over the top for these kinds of limits. Try proving that, for any rational limit (quotient of polynomials):
$$\lim_{n\to\infty} \frac{a_qn^q+a_{q-1}n^{q-1}+\dots+a_1n+a_0}{b_pn^p+b_{p-1}n^{p-1}+\dots+b_1n+b_0}=\left\{\begin{array} {r} 0,  \quad p>q \\ \infty, \quad p<q \\ \frac{a_q}{b_p}, \quad p=q \end{array}\right.$$
