$$\lim_{n\to+\infty}\frac{11^n n^4 + 9^n n^9} {7^{2n} +1}=0$$
I have used the fact exponentials grow faster than polynomials for this question but was wondering whether there was an algebraic way e.g. with the ratio lemma.
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3That is probably simplest. Do observe that $7^2$ outguns both $11$ and $9$. – Jyrki Lahtonen Oct 30 '20 at 09:15
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1$\frac{11^n n^4+9^n n^9}{7^{2n}+1}<2\frac{11^n n^9}{7^{2n}}$ – PNDas Oct 30 '20 at 09:19
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Use ratio test with PNDas' hint to prove the bigger sequence converges to 0. And hence the smaller sequence also converges to 0. – Ameet Sharma Oct 30 '20 at 11:10
4 Answers
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According to the suggestion by Jyrki we have
$$\frac{11^n n^4 + 9^n n^9} {7^{2n} +1} \le \frac{11^n n^9 + 11^n n^9} {7^{2n} }=2\cdot\frac{ 11^n n^9} {49^n }=2\cdot\frac{ n^9} {\left(\frac{49}{11}\right)^n }\le 2 \cdot \frac{n^9}{4^n}$$
Refer also to the related
user
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so then do you use the fact that 4^n will grow at a rate much greater than n^9? – Oct 30 '20 at 09:32
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@Emily.KH Yes you can take this a general fact which can be proved in many ways, ratio test works or as an laternative by induction, depending on what are you given to know and allowed to use. I add some reference for that. – user Oct 30 '20 at 09:34
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From your expression, $7^{2n}= 49^n$ seems to be the dominant term. We can divide both numerator and denominator by $49^n$:
$$a_n = \frac{n^4 \left(\frac{11}{49} \right)^n + n^9\left(\frac{9}{49} \right)^n }{1 + \left(\frac{1}{49}\right)^n}= \frac{\dfrac{n^4}{\left(\frac{49}{11} \right)^n}+ \dfrac{n^9}{\left(\frac{49}{9} \right)^n}}{1 + \left(\dfrac{1}{49} \right)^n}$$
Since $\left(\left(\frac{1}{49} \right)^n\right)$ is a basic null sequence and by the reciprocal rule, $\left(\left(\frac{49}{11} \right)^n\right), \left(\left(\frac{49}{9} \right)^n\right)$ both tend to infinity which makes $\left(\dfrac{n^4}{\left(\frac{49}{11}\right)^n}\right)$ and $\left(\dfrac{n^9}{\left(\frac{49}{9}\right)^n}\right)$ both basic null sequences. $$\lim_{n \to \infty} a_n = \frac{0 + 0}{1 + 0} = 0$$
Alternatively, you can also say that $\left(n^4\left(\frac{11}{49} \right)^n\right)$ and $\left(n^9\left(\frac{9}{49} \right)^n\right)$ are basic null sequences.
John_dydx
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This argument is wrong. The terms in the numerator are indeterminate forms $\infty\cdot0$ and you cannot conclude so expeditiously. And if you wanted to justify, you would probably invoke "exponentials grow faster than polynomials". So no progress on the question. – Oct 30 '20 at 11:41
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1You can adjust your previous answer in this way $$a_n = \frac{n^4\left(\frac{11}{49}\right)^n + n^9\left(\frac{9}{49}\right)^n}{1 + \left(\frac{1}{49}\right)^n}=\frac{\frac{n^4}{\left(\frac{49}{11}\right)^n} + \frac{n^9}{\left(\frac{49}{9}\right)^n}}{1 + \left(\frac{1}{49}\right)^n}=\ldots$$ – user Oct 30 '20 at 12:30
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Do some basic asymptotic analysis:
It is easy to check that $9^nn^9=o\bigl(11^nn^4\bigr)$, so the numerator is asymptotically equivalent to $11^nn^4$ and $$\frac{9^nn^9+11^nn^4}{7^{2n}+1}\sim_\infty\frac{11^nn^4}{7^{2n}}=\Bigl(\frac{11}{49}\Bigr)^n n^4\to 0.$$
Bernard
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Consider the sequence
$$s_n:=\frac{n^k}{a^n}$$ where $a>1$.
We have
$$\frac{s_{n+1}}{s_n}=\left(\frac{n+1}n\right)^k\frac1a.$$
Now if we take
$$n>\frac1{\sqrt[k]a-1},$$ we have $$\frac{s_{n+1}}{s_n}<\left(\frac{n+1}n\right)^k\frac1a=r<1$$ and by induction
$$s_{n+m}<s_nr^m\to0.$$
This proves that any exponential grows faster than any power and it is not big deal to generalize to any polynomial.
E.g. with
$$\frac{n^9}{7^{2n}},$$ as of $n\ge2$ the sequence is dominated by a geometric series of common ratio $\dfrac{19683}{25088}.$