I would like to prove if the following sequence is inc. or decrea. and then if it is bounded.
So this is the sequence:$$\lim_{n \to \infty}\frac{2n}{n^2+1}$$.
Now I really don’t know how to prove this.But this is how I would go about this problem.So we know that if a sequence converges then it’s bounded.The limit of this thing is 0, therefore it’s bounded.But how do I evaluate the bounds?I would take the derivative which would come out to be increasing for -1<x<1.Therefore we can conclude that 1 is the least upper bound of this sequence.I don’t really know if this is correct, so please point out any logical mistake.The lower bound is clearly 0(numerator and denominator can just be positive).But I have also seen some proofs by induction.Could someone show me the whole proof by induction?I have seen proofs when the sequence is strictly increasing or decreasing but in this case the behaviour of the base case isn’t obvious.Can we say $$a_{1} > a_{0}$$ and then conclude the opposite(that the function is not monotonous)?Thank you for checking my work and and attempting to clear my doubts.
Best Answer
Notice that $$0\leq \frac{2n}{n^2+1}\leq \frac{2n}{n^2}=\frac{2}{n}<2$$
Hence $a_n=\frac{2n}{n^2+1}$ is bounded by $2$
For show that $a_n$ is increasing you should prove that $$a_{n}<a_{n+1}$$
it is $$ \frac{2n}{n^2+1}< \frac{2(n+1)}{(n+1)^2+1}$$