An unorthodox way of converting a recurrent sequence into a non-recurrent one

Question

The problem is:

A sequence is given with: $$ a_1 = 2016, a_{n+1} = \sqrt{a_n^2 + 4 + \frac{1}{a_n^{222}}}$$ Find: $$ (a) \quad \lim_{n \to \infty}{a_n} \quad (b) \lim_{n \to \infty}{\frac{a_n^2}{n}} \quad (c) \lim_{n \to \infty}{\frac{\sum_{k=1}^{n}{a_k}}{n \sqrt{n}}}$$

The (a) part I was able to solve by proving that the sequence is increasing and not bounded by above. Therefore $$ \lim_{n \to \infty}a_n = +\infty$$

Then for (b) I applied the Stolz theorem: $$ \lim_{n \to \infty}{\frac{a_n^2}{n}} = \lim_{n \to \infty}{\frac{a_{n+1}^2 - a_n^2}{n+1-n}} = \lim_{n \to \infty}{(a_n^2 + 4 + \frac{1}{a_n^{222}} - a_n^2)} = \lim_{n \to \infty}{(4 + \frac{1}{a_n^{222}})} = 4$$

But the (c) part is where I get confused. I've tried applying the Stolz theorem there as well but no luck since the n's don't cancel out like they did in (b).

I did have one thing on my mind though but I'm not sure how formally correct it is. Let's look at the first few sequence members: $$a_1 = 2016, a_2 = \sqrt{2016^2 + 4 + \frac{1}{2016^{222}}}, a_3 = \sqrt{2016^2 + 4 + \frac{1}{2016^{222}}+ 4 + \frac{1}{a_2^{222}}} = \sqrt{2016^2 + 8 + \frac{1}{2016^{222}} + \frac{1}{a_2^{222}}}$$

And now as we keep going we are going to get a new $$ \frac{1}{a_n^{222}}$$ member inside for each n. So that still makes this sequence recursive. But can I simply disregard that part since it obviously always converges to zero, not matter what the value of n is? Or maybe I could say something like

$$ \text{Let } \frac{1}{a_n^{222}} = Z(n) \quad \text{ Then } \lim_{n \to \infty}Z(n) = 0 \quad (\forall n)$$

And now I rewrite my sequence as:

$$ a_n = \sqrt{(2016)^2 + 4(n-1) + \sum_{i=1}^{n-1}Z(i)} \quad \forall n \geq 2$$

Now let's finally look at the (c) problem:

$$\lim_{n \to \infty}{\frac{\sum_{k=1}^{n}{a_k}}{n \sqrt{n}}} = \lim_{n \to \infty}{\frac{a_1 + a_2 + ... +a_n}{n \sqrt{n}}} = \lim_{n \to \infty}{\frac{2016 + \sqrt{2016^2 + 4 + Z(1)} + ... + \sqrt{2016^2 + 4(n-1) + \sum{Z(n})} }{n \sqrt{n}}}$$

And now I apply Stolz theorem:

$$ = \lim_{n \to \infty}{\frac{\sqrt{2016^2 + 4n + \sum{Z(n+1)}}}{(n+1)\sqrt{n+1} - n\sqrt{n}}} = 0 $$

Since the top degree of $n$ is $\frac{1}{2}$ and the bottom one is $ \frac{3}{2}$.

So can anyone tell me whether this solutions is correct? If it isn't, how would you solve it? If it is correct but my process is wrong I'd also like to know where I went wrong.

Thanks in advance.

Answer 1

For $a$ we can go like this:

The sequence is clearly increasing, it's also positive, we just need to prove that the limit is infinite. If it were finite and equal to $g\geq a_1>0$, we would have $$g=\sqrt{g^2+4+\frac{1}{g^{222}}}\implies \frac{1}{g^{222}}=-4 $$ which is impossible, so that $\lim_{n\to\infty} a_n = \infty$.

Your solution for $b$ looks good.

For $c$:

$$\lim_{n\to\infty} \frac{\sum\limits_{k=1}^n a_k}{n^{3/2}} = \lim_{n\to\infty} \frac{a_{n+1}}{\sqrt{n+1}+\frac{n}{\sqrt{n+1}+\sqrt{n}}} = \lim_{n\to\infty} \frac{a_n}{(3/2)\sqrt{n}} = \frac{2}{3}\cdot2 = \frac{4}{3} $$ First equality is Stolz's theorem, $$(n+1)^{3/2}-n^{3/2} = n(n+1)^{1/2}+(n+1)^{1/2}-n^{3/2}=\\=n(\sqrt{n+1}-\sqrt{n})+\sqrt{n+1}=\frac{n}{\sqrt{n+1}+\sqrt{n}}+\sqrt{n+1},$$ for the second we used that $\frac{\frac{n}{\sqrt{n+1}+\sqrt{n}}+\sqrt{n+1}}{\sqrt{n+1}}\to 3/2$ as $n\to\infty$, third we calculated earlier.

About your approach, the problem was that bottom part behaves like $n^{1/2}$, not like $n^{3/2}$. $$\lim_{n\to\infty}\frac{\sqrt{2016^2+4n+\sum_{k=1}^n Z(k)}}{(n+1)\sqrt{n+1}-n\sqrt{n}} = \lim_{n\to\infty} \frac{\sqrt{2016^2+4n+\sum_{k=1}^n Z(k)}}{(3/2)\sqrt{n}}=\\= \lim_{n\to\infty} \frac{2}{3}\sqrt{\frac{2016^2}{n}+4+\frac{\sum_{k=1}^n Z(k)}{n}}$$ But $$\lim_{n\to\infty} \frac{\sum_{k=1}^n Z(k)}{n} = \lim_{n\to\infty} Z(n) = 0 $$ So that $$\lim_{n\to\infty} \frac{2}{3}\sqrt{\frac{2016^2}{n}+4+\frac{\sum_{k=1}^n Z(k)}{n}} = \frac{2}{3}\cdot\sqrt{4} = \frac{4}{3}. $$