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I have to find and prove the limit of the sequence $X_n=$ $\frac {n^{100}}{1.01^n}$

What is the easier way?

I tried to use Bernoulli's inequality to say lim$\frac {n^{100}}{1.01^n}$ = lim$\frac {n^{100}}{(1+1/10)^n}$ and $ (1+1/10)^n \geq 1+n(1/10).$ But I could get anything. I think another way is to use the Squeeze Theorem but I have not could find the correct sequences.

Any Ideas?

I only can use the definition, Squeeze Theorem, the Bernoulli's inequality or using operations to reduce the sequence.

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  • $\begingroup$ You now have five people who have answered your question. Do any of the answers provided successfully answer your question? If so, (this goes for all the other questions you've asked on MSE as well) you should click the check mark next to the answer that you feel best answered your question. This will let other MSE users know that your question has an answer, and it will give some points to the user who provided the answer. $\endgroup$ – graydad Oct 21 '15 at 5:16
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    $\begingroup$ You accepted the only "wrong" answer :) Well, it's not wrong but it doesn't give you the limit of the sequence. $\endgroup$ – lcv Nov 13 '15 at 6:23
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We have $$\frac {n^{100}}{1.01^{n}} = \left(\frac{n}{1.01^{n/100}}\right)^{100} $$ By Bernoulli's inequality on the inside quantity, $$\begin{align}\left(\frac{n}{\left(1+\frac{1}{100}\right)^{n/100}}\right)^{100} &\leq \left(\frac{n}{1 + \frac{1}{100}\cdot\frac{n}{100}}\right)^{100} \\ &< \left(\frac{n}{\frac{n}{10000}}\right)^{100} \\ &= 10000^{100} \end{align}$$

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Theorem. If $a,b > 0$, then $(\log x)^{a}/x^{b} \to 0$ as $x \to \infty$.

Corollary. If $a,b > 0$, then $x^{a}/e^{xb} \to 0$ as $x \to \infty$.

The corollary follows immediately from the fact that formally we have $x^{a}/e^{xb} = (\log t)^{a}/t^{b}$ and the theorem.

Take $a := 100$ and $b:= \log (1.01)$; then the statement that you want to prove follows directly.

To prove the theorem, just note that, given any $a,b > 0$, we have $x \geq 1$ only if for all $c > 0$ we have $$ \frac{(\log x)^{a}}{x^{b}} = \frac{\big( \int_{t=1}^{x}\frac{1}{t} \big)^{a}}{x^{b}} \leq \frac{\big( \int_{t=1}^{x} t^{c-1} \big)^{a}}{x^{b}} = \frac{(x^{c} - 1)^{a}}{c^{a}x^{b}} < \frac{x^{ac}}{c^{a}x^{b}}; $$ then take $c := \frac{b-1}{a}$.

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When you are dealing with sequences it is better to use a theorem related to sequences. We will use the following result:

Theorem: If $0 < r < 1$ then $r^{n} \to 0$ as $n \to \infty$ and if $r > 1$ then $r^{n} \to \infty$ as $n \to \infty$.

Also it is best to analyze the sequence $a_{n} = n^{k}/a^{n}$ in general where $k, a$ are positive real numbers. For the current question we have $a = 1.01$ and $k = 100$.

We have $$\frac{a_{n + 1}}{a_{n}} = \frac{1}{a}\left(\frac{n + 1}{n}\right)^{k} \to a\cdot 1^{k} = \frac{1}{a}$$ as $n \to \infty$. We will show that if $0 < a \leq 1$ then $a_{n} \to \infty$ and if $a > 1$ then $a_{n} \to 0$ as $n \to \infty$. Note that when $a = 1$ then $a_{n} = n^{k}$ so that $a_{n} \to \infty$ as $n \to \infty$.

Let's start first with the case $a > 1$. Then clearly $b = 1/a < 1$ and therefore there is a number $r$ such that $b < r < 1$. Since $a_{n + 1}/a_{n} \to b$ as $n \to \infty$, it follows that there is a positive integer $m$ such that $a_{n + 1}/a_{n} < r$ for all $n \geq m$. Hence we can see that $$0 < \frac{a_{m + n}}{a_{m}} = \frac{a_{m + 1}}{a_{m}}\cdot\frac{a_{m + 2}}{a_{m + 1}}\cdots\frac{a_{m + 1n}}{a_{m + n - 1}} < r^{n}$$ Since $0 < r < 1$ it follows that $r^{n} \to 0$ as $n \to \infty$ and hence applying Squeeze theorem to the above equation we see that $$a_{m + n}/a_{m} \to 0$$ and therefore $a_{m + n} \to 0$. It follows that $a_{n} \to 0$. For the current question we have $a = 1.01$ so that $a > 1$ and hence it is covered by this case and the desired limit is $0$.

In exactly the same way we can show that if $0 < a < 1$ then $a_{n} \to \infty$. For this case we have $b = 1/a > 1$ and we need to choose $1 < r < b$ and use the fact that $a_{n + 1}/a_{n} > r$ for all large values of $n$. The result will follow because in this case $r^{n} \to \infty$.


The above proof is pretty standard (and should be available in any textbook which deals with limit of sequences) and its beauty is that it does not make use of exponential and logarithm functions (and no differentiation/L'Hospital's Rule)

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Since $$ e^x> \frac{x^{101}}{101!} \quad \forall x \ge 0, $$ we have $$ 1.01^n=e^{n\ln1.01} >\frac{n^{101}\ln1.01}{101!} \quad \forall n\ge 1. $$ Therefore $$ 0<\frac{n^{100}}{1.01^n}<\frac{101!}{\ln1.01}\cdot\frac1n \quad \forall n\ge 1. $$ It follows that $$ 0\le\lim_{n\to\infty}\frac{n^{100}}{1.01^n}\le \lim_{n\to\infty}\frac{101!}{\ln1.01}\cdot\frac1n=0, $$ i.e. $$ \lim_{n\to\infty}X_n=0. $$

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We will show that $\displaystyle \lim_{n\to\infty}\frac{n^k}{1.01^n}=0$ for all $k$. In particular, $k=100$ is the sequence you are asking about.

When $k=0$, we have $\lim_{n\to\infty}\frac{1}{1.01^n}$, which is $0$ since $1.01^n$ increases without bound.

Suppose that $k>0$ and that $\displaystyle \lim_{n\to\infty}\frac{n^{k-1}}{1.01^n}=0$. Since $\displaystyle \lim_{n\to\infty}\frac{n^k}{1.01^n}$ is an $\frac{\infty}{\infty}$ indeterminate form, we may apply l'Hopital's rule to see that the limit is equal to $\displaystyle \lim_{n\to\infty}\frac{kn^{k-1}}{\ln(1.01)1.01^n}=\frac{k}{\ln 1.01}\lim_{n\to\infty}\frac{n^{k-1}}{1.01^n}$, which equals zero by hypothesis.

By induction, the limit is zero for all $k\geq 0$.

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  • $\begingroup$ L'Hospital's is not allowed for this problem $\endgroup$ – graydad Oct 21 '15 at 3:23

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