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?
Best Answer
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.