Showing a recursively defined sequence is convergent For $a_1=1$ and $a_{n+1} = 1 + \frac{a_n}{3+n}$, I want to show that the sequence $a_n$ converges. I will use the Monotone Convergence Theorem.
Of course, the sequence is bounded below by $1$.
Now I want to show that the sequence is (except for a first few terms) decreasing.
The difference between consecutive terms is:
$$a_{n+1}-a_n = 1 + \frac{a_n}{3+n}-a_n = 1 - \frac{2a_n+na_n}{3+n}$$
So, we want to show that $$\frac{2a_n+na_n}{3+n} = a_n \frac{2+n}{3+n} \geq 1$$ for $n$ big enough.
How can I proceed?
Alternative ways are also welcome.

We have $$a_n=1+\frac{a_{n-1}}{3+(n-1)}\ge 1+\frac{1}{3+(n-1)} =\frac{3+n}{2+n}.$$
