How to show $a_n = \frac{n+1}{n-1}$ is strictly decreasing by induction?

Question

I can show this fact otherwise, but I can't seem to figure out the simple algebra to prove it by induction...Could someone provide a hint? I just need a push in the right direction. It seems too simple to be this difficult.

(Please don't give me the answer, just a hint will do).

Answer 1

This is the easy way. Write

$$a_n = \frac{n+1}{n-1} = \frac{n-1 + 2}{n-1} = 1 + \frac{2}{n-1}.$$

Then use induction to show that $2/(n-1)$ is decreasing and the hence $a_n$ is decreasing.

Answer 2

Proof by induction is as follows:

Base step: Show the statement is true for the smallest value of n that matters. In your case, I'm guessing the smallest n is 2?

Inductive Step: show that if the statement is true for some n then it is true for n+1.

By induction, your proof is complete.

It's easy to show that it's true for the base step. $a_2=3$. $a_3=\frac{4}{2}=2$. And $2<3$.

Can you do the inductive step?

Answer 3

$\begin{array}\\ a_n-a_{n+1} &=\frac{n+1}{n-1}-\frac{n+2}{n}\\ &=\frac{(n+1)n-(n+2)(n-1)}{(n-1)n}\\ &=\frac{n^2+n-(n^2+n-2}{(n-1)n}\\ &=\frac{2}{(n-1)n}\\ &> 0\\ \end{array} $

Answer 4

Not a proof by induction but it never hurs to notice

$$a_n = \frac{n+1}{n-1}$$ $$a_{n+1} = \frac{(n+1)+1}{(n+1)-1}= \frac{n+2}{n}$$

Notice that $$n^2+n-2 \lt n^2+n$$ $$(n+2)(n-1) \lt n(n+1)$$

$$\frac{n+2}{n} \lt\frac{n+1}{n-1}$$ $$a_{n+1}\lt a_n$$