Use formal definition of a limit of a sequence to prove the following:

Question

$$\lim_{n\rightarrow\infty} \frac{n+1}{n+2}=1$$

$$\lim_{n\rightarrow\infty} \frac{n^{3}-1}{n^{4}-1} c) lim $n!/n^{n}=0$ as n approaches inifinity

I tried proving those but I'm not sure if i did it right so would appreciate if anyone could point out if i missed any steps or if I can improve them. Thank you

a) |[(n+1)/(n+2)]-1| = |[(n+1)-(n+2)]/(n+2)| = 3/(n+2)

n+2>n

1/(n+2)<1/n

3/(n+2)<3/n

So for any epsilon>0, 3/epsilon>0

By Archimedean property, there exists N in natural numbers such that N>3/epsilon

When n≥N, n≥N>3/epsilon

which implies that 1/n≤1/N<1/(3/epsilon) implying 3/n≤3/N<3/(3/epsilon)

So, $|x_{n}-L|=3/(n+2)$<3/n≤3/N<3/(3/epsilon)=epsilon

Therefore, $|x_{n}-L|$=3/(n+2)<epsilon

implying that $|x_{n}-L|$<epsilon

Then by definition of limit, $x_{n}$=(n+1)/(n+2) converges to real number and its limit is 0.

b)for any n≤N in natural numbers, $n^{3}-1<n^{3}$ for any n>2, we have 0<$n^{4}-n^{3}$<$n^{4}-1$

so 1/($n^{4}-1$)<1/($n^{4}-n^{3}$). Hence, for n>2, 0<($n^{3}-1$)/($n^{4}-1$)<$n^{3}$/($n^{4}-1)$<$n^{3}$/($n^{4}-n^{3}$)=1/(n-1)

So given any epsilon>0, it'll be sufficient to choose N such that 1/(N-1)<epsilon, so that N>((1/epsilon)+1)

Given epsilon>0, choose N=(1/epsilon)+1. Then N>(1/epsilon)+1, hence 1/(N-1)<epsilon. Moreover, since N>2, given any n≥N, we have 0<$(n^{3}-1)/(n^{4}-1)$<$n^{3}/(n^{4}-1)$<$n^{3}/(n^{4}-n^{3})$=1/(n-1)<1/(N-1)<epsilon

Therefore, the stated limit is indeed 0.

c) |n!/$n^{n}$|<1/n for all n in Natural numbers excluding 0

for any epsilon>0, 1/epsilon>0, there exists N in Natural numbers such that N>1/epsilon, when n≥N, n≥N>1/epsilon

implying that 1/n≤1/N<1/(1/epsilon), therefore, 1/n≤1/N<epsilon

So, |$x_{n}-L$|=n!/$n^{n}$<1/n≤1/N<epsilon

Therefore, |$x_{n}-L$|<epsilon

By definition of a limit, $x_{n}=n!/n^{n}$ converges to real number and its limit is 0.

Answer 1

Yes, they seem decent to me.

good job, (Even through it was a bit of a pain to read this as you didn't typeset any maths)

Answer 2

)b and c) look good. In a), however, you made a mistake in your first line. This compromises the rest of your proof (it needs correction too), but your reasoning is correct. For $n \in \mathbb{N}, (n+1) - (n+2) = n + 1 - n - 2 = -1$, which makes sense as $(n+1) / (n+2) < 1$ (so the expression inside the absolute value is negative).

Instead, you should get: $$ \bigg\lvert \frac{(n+1) - (n+2)}{(n+2)}\bigg\rvert = \frac{1}{n+2}. $$ You want to show that this converges to $0$. Given $\epsilon > 0$, by the Archimedean property there is an integer $N$ such that $1/\epsilon < N$, so $1/\epsilon < n < n+2$ for all $n \ge N$. But then for all integers $n \ge N$, $$ \frac{1}{n + 2} < \epsilon, $$ which is what you needed to show.