Question

(1 pt) Let an=n+1n+3. Find the smallest number M such that: (a) |an−1|≤0.001 for n≥M M= 1997 (b) |an−1|≤0.00001 for n≥M M= 200003 (c) Now use the limit definition to prove that limn→[infinity]an=1. That is, find the smallest value of M (in terms of t) such that |an−1|M. (Note that we are using t instead of ϵ in the definition in order to allow you to enter your answer more easily). M= 2/t - 3t (Enter your answer as a function of t)

Answer 1

We want to prove

$\displaystyle\lim_{n\to\infty}\frac{n+1}{n+3}=1$

which is akin to saying that, for any given $\varepsilon>0$, we guarantee that

$\left|\dfrac{n+1}{n+3}-1\right|$

for all $n$ exceeding some threshold $N=N(\varepsilon)$.

We want to end up with

$\left|\dfrac{n+1}{n+3}-1\right|=\left|-\dfrac2{n+3}\right|=\dfrac2{n+3}$

$\implies\dfrac{n+3}2>\dfrac1\varepsilon\implies n>\dfrac2\varepsilon-3$

which suggests that we guarantee that $a_n$ is arbitrarily close to 1 if $N=\left\lceil\dfrac2\varepsilon-3\right\rceil$. (Take the ceiling to ensure $N$ is a natural number.)

Now for the proof: Let $\varepsilon>0$ be given. Then for $n>N=\left\lceil\dfrac2\varepsilon-3\right\rceil$ we have

$n>\left\lceil\dfrac2\varepsilon-3\right\rceil\le\dfrac2\varepsilon-3\implies\dfrac2{n+3}$

QED