I've the following sequence: $b_{n+1}=b_{n}-\frac{1}{2n+2}+\frac{1}{2n+3}$ with $b_{1}=\frac{5}{6}$. I need to show this is a monotonic decreasing sequence, so $b_{n}>b_{n+1}$. I think I should use induction, thus:
basis: $b_{1}=\frac{5}{6}$ and $b_{2}=\frac{47}{60}$ thus $b_{1}>b_{2}$, it holds.
hypothesis: if $b_{n}>b_{n+1}$ then it follows that $b_{n+2}>b_{n+1}$.
Proof: Assume $b_{n}>b_{n+1}$. And this is where I'm stuck. How do I get from here to $b_{n+2}>b_{n+1}$?
Thanks in advance...
On
$b_{n+1}-b_n=\dfrac{-1}{2n+2}+\dfrac{1}{2n+3}=\dfrac{-(2n+3)+(2n+2)}{(2n+2)(2n+3)}=\dfrac{-1}{(2n+2)(2n+3)}<0$
So $b_{n+1}<b_n$, you just need to add the two fractions, numerator is negative and denominator is positive, so overall a negative number.
There is no need for induction, since this is valid for any $n$.
No need to use induction just write out $b_{n+1}<b_n$ which is equivalent with $$b_n-\frac1{2n+2}+\frac1{2n+3}<b_n\\\frac{1}{2n+2}>\frac{1}{2n+3}$$ Which is true.