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For some sequence $x_n > 0$, we know that sequence $x_n^2 - 2x_n - 15 +4^{-n}$ converges.

Does this mean that $x_n$ also converges?

If so, proof that it is convergent, otherwise provide a counterexample.

First of all, let's define $y_n = x_n^2 - 2x_n - 15 +4^{-n}$, as $n \to \infty \Rightarrow y_n = x_n^2 - 2x_n - 15 $.

Assume $\lim y_n = a$, then $a = x_n^2 - 2x_n - 15$. This equation gives two roots: $x_n = 1 + \sqrt{4a + 16}$ and $x_n = 1 - \sqrt{4a + 16}$. And we know that both of them should be $>0$, since $x_n > 0$.

From $1 - \sqrt{4a + 16} > 0$, we get $-4 \leq a < \frac{-15}{4}$.

So $y_n$ converges, when $x_n$ is either $1 - \sqrt{4a + 16}$ or $1 + \sqrt{4a + 16}$. So I can take $x_n = \{ 1 - \sqrt{4a + 16}, 1 + \sqrt{4a + 16}, 1 - \sqrt{4a + 16}, 1 + \sqrt{4a + 16}, \dotso \}$, so $y_n$ still converges, while $x_n$ is not.

Also if $1 - \sqrt{4a + 16} = 1 + \sqrt{4a + 16}$, then $x_n$ also converges.

Despite all my thoughts, I'm pretty sure that this solution have some mistakes, for example, I'm not sure that I can cancel out $4^{-n}$.

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Hint: We can write $x_n^2-2x_n-15+4^{-n} = (x_n-1)^2-16+4^{-n}$.

Now, consider the sequence $x_n = 1+\dfrac{1}{2}(-1)^n$.

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    $\begingroup$ @SimpleArt $x_n > 0$ $\endgroup$ – user384138 Dec 29 '16 at 0:52

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