Proving a recursion is less than two

Question

The sequence $(a_n)$ is defined by $a_1 = \frac{1}{2}$ and $a_n = a_{n - 1}^2 + a_{n - 1}$ for $n \ge 2.$

Prove that $\frac{1}{a_1 + 1} + \frac{1}{a_2 + 1} + \dots + \frac{1}{a_n + 1} < 2$ for all $n \ge 1.$

So far, I have manipulated the given recursion into $a_n = a_{n - 1} (a_{n - 1} + 1)$. But now I am stuck.

Hint: $a_{k+1} = a_k(a_k+1) \implies \frac{1}{a_{k+1}} = \frac{1}{a_k(a_k+1)} = \frac{1}{a_k} - \frac{1}{a_k+1} \iff \frac{1}{a_k+1} = \frac{1}{a_k} - \frac{1}{a_{k+1}} $ — achille hui, Jun 04 '20 at 13:43

score 2 · Answer 1 · answered Jun 04 '20 at 13:47

We claim a stronger result; namely that $\frac{1}{a_1+1} + \cdots + \frac{1}{a_{n-1}+1} + \frac{1}{a_n}<2$, for all $n$. This obviously holds for small $n$; you can check this yourself.

Observe that $$\frac{1}{a_n} = \frac{1}{(a_{n+1}+1)a_{n+1}} = \frac{1}{a_n+1} + \frac{1}{a_{n+1}}.$$

The claim follows by induction.

score 1 · Accepted Answer · answered Jun 04 '20 at 14:05

1

After all we have by @hdighfan solution: $$\frac{1}{a_n~+1}=\frac{1}{a_n}-\frac{1}{a_{n+1}}$$ By telescopic summing we have $$S_n=\sum_{k=1}^n \frac{1}{a_k~+1}=\frac{1}{a_1}-\frac{1}{a_{n+1}}<2,$$ as $a_1=1/2$.

answered Jun 04 '20 at 14:05

Z Ahmed

43,235

Thank you so much! I wish i could upvote you and @hdighfan – fds-mqdwqmqkdwl Jun 04 '20 at 14:08
@fds-mqdwqmqkdwl you are welcome. – Z Ahmed Jun 04 '20 at 15:56

Proving a recursion is less than two

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