[College Calc 01-02. Limits of Sequences] #12

Let us examine for what value of q does the subsequence $\{a_{m}\}$ where $m$ is an odd natural number converges.

$\displaystyle \lim_{m\to\infty} a_{m} = \lim_{m\to\infty} qm^{2} - \sqrt{4m^{4}+pm^{2}+8m} =\lim_{m\to\infty} \frac{q-\sqrt{4+\frac{p}{m^{2}}+\frac{8}{m^{3}}}}{\frac{1}{m^{2}}} = \lim_{h\to 0^{+}} \frac{q-\sqrt{4+ph^{2}+8h^{3}}}{h^{2}}$ . Hence $q$ must be equal to $2$ to maintain the $\frac{0}{0}$ form otherwise it diverges.

hence applying L'Hopital's Rule , This limit evaluates to $\displaystyle \lim_{h\to 0^{+}} \frac{2ph}{2\cdot 2h\sqrt{4+ph^{2}+8h^{3}}}=\frac{p}{4}$ .

But for this value of $q$ . The subsequence of even terms is unbounded above as it is $\,\,\,\,\,\,\infty +\infty\,\,\,\,\,\,\,\,$ form and tends to $\infty$ as $n\to \infty$ . Hence $\lim \text{sup}\{a_{n}\} = \infty$ . Hence we have to reject this value of $q$

So , Let us examine the value of q for which the subsequence of even terms $\{a_{k}\}$ where $k$ is an even natural number converges.

$\displaystyle \lim_{k\to\infty} a_{k}= \lim_{k\to\infty} qk^{2}+\sqrt{4k^{4}-pk^{2}+8k} = \lim_{h\to 0^{+}} \frac{q+\sqrt{4-ph^{2}+8h^{3}}}{h^{2}}$ .

For convergence $q$ must equal $-2$ to maintain $\frac{0}{0}$ form.

Hence again by L'Hopital's Rule , we have the limit is

$\displaystyle \lim_{h\to 0^{+}} \frac{-2ph}{2\cdot 2h\sqrt{4-ph^{2}+8h^{3}}}=\frac{-p}{4}$ .

For this value of $q$ the $\displaystyle \lim \text{inf}\{a_{n}\} = -\infty$ as the subsequence of odd terms is of $\,\,\,\,\,\,\,\,\,-\infty -\infty\,\,\,\,\,\,\,\,$ form hence it is unbounded below and tends to $-\infty$

For $\displaystyle \lim \text{sup}\{a_{n}\} = 1$ we must have $\displaystyle p=-4$ .

Hence for $q=-2$ and $p=-4$ we have our required condition.

Suppose $q>-2.$ Then the supremum of the entire sequence, $\sup\{a_n\},$ is positive infinity, which is a contradiction.

$q<-2,$ on the other hand, yields a contradiction in that $\lim_{n\to\infty}\sup\{a_m, a_{m+1}, \cdots\}=\infty.$

Thus we know we must have $q=-2.$

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As $n$ gets larger, the $n^2$ -term inside the square root becomes more and more negligible. Thus the supremum is decided by the subsequence $a_{2n},$ for which the $(-1)^n$ in front of the square root would be positive, yielding a larger result.

$a_{2n}=-8n^2+\sqrt{64n^4-4pn^2+16n},$

and taking the limit we have

$\lim_{n\to\infty} a_{2n}=\lim_{n\to\infty}\frac{-4pn^2+16n}{16n^2}=-\frac{p}{4},$

which must be equal to $1.$ Thus $p=-4,$ so that $p+q=\boxed{-6}.$