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.

  • 1
    $\begingroup$ Normally I approach these problems by showing $a_{n+1}\le a_n$ by induction to show that the sequence is monotonically decreasing. $\endgroup$ – Dave Apr 25 '16 at 19:31
  • $\begingroup$ @Dave Isn't that exactly what OP tries? $\endgroup$ – Hagen von Eitzen Apr 25 '16 at 19:32

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


Note that

$$ a_{n}\frac{2+n}{3+n}\geq 1\iff a_{n}\geq\frac{3+n}{2+n}. $$

Since $a_{1}=1$, an easy induction shows that $a_{n}\geq 1$ for all $n$. Thus, your desired inequality follows.


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