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Prove that the sequence $\left(x^n\right)_{n\in\mathbb{N}}$ diverges when $x>1$

My attempt:

Let $L$ be a real number. I will show that $x^n$ does not converge to $L$ for all $x>1$. By definition of limits of a sequence, if a sequence $(a_n)$ converges to $L$ we can write $\lim_\limits{n\to\infty}a_n = L$. If a sequence is not converging to any real number $L$ then we say that the sequence is divergent and $\lim_\limits{n\to\infty}a_n$ is undefined.

Let's assume, for contradiction, that $\lim_\limits{n\to\infty}x^n = L$.

Now I am stuck...

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  • $\begingroup$ Mathematical formulae look better in $\LaTeX$. Here is a quick tutorial. $\endgroup$ – Τίμων Apr 8 '16 at 20:54
  • $\begingroup$ @AndréNicolas Your comment appeared while I was writing my answer. $\endgroup$ – egreg Apr 8 '16 at 21:06
  • $\begingroup$ Your answer makes the comment unnecessary! $\endgroup$ – André Nicolas Apr 8 '16 at 21:17
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Recall the Bernoulli inequality $$ (1+t)^n\ge 1+nt $$ for $t>-1$. If you set $t=x-1>0$, you have $$ x^n\ge 1+n(x-1) $$ Fix $M>0$. Then the inequality $$ 1+n(x-1)>M $$ is satisfied for $n>(M-1)/(x-1)$. Therefore, for $n>(M-1)/(x-1)$ you have $$ x^n>M $$ and the sequence diverges.

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Well if we have the series $$\sum_{n=1}^{\infty}a_n$$ Then the divergence test says if $$\lim_{n\to\infty}a_n\neq 0$$ Then the series diverges. Applying this is the series at hand, $$\lim_{n\to\infty}x^n=\infty$$ if $x>1$, is enough to prove what you are interested in. And for $x=1$ just do a simple comparison test to the harmonic series.

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  • $\begingroup$ We can't prove it with series because it hasn't been introduced yet $\endgroup$ – combo student Apr 8 '16 at 21:00
  • $\begingroup$ Then still apply the limit and it will diverge. $\endgroup$ – Will Fisher Apr 8 '16 at 21:01
  • $\begingroup$ I also don't see what any of this has to do with the limit of the OP's sequence. $\endgroup$ – pjs36 Apr 8 '16 at 21:08
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Rewrite it this way :

$L=\lim_\infty x^n=\lim_\infty x^{2n}=(\lim_\infty x^n)^2=L^2$

Hence $L=0$ or $L=1$ which is impossible under your assumption that $x>1$.

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We know that the sequence converges only if it is Cauchy. Now for an arbitrary $k\in \mathbb{N}$ consider the difference: $$x^{k+1}-x^{k}=x^k(x-1)$$ For $x>1$ we obviously have $x^k>1$ and $x-1=\epsilon_x>0$.

Thus $\forall M\in \mathbb{N}, n>M\implies x^{n+1}-x^{n} \geq \epsilon_x$

So $(x_n)$ is not Cauchy and so is not convergent. Furthermore $(x_n)$ is strictly increasing, and so it must tend to infinity.

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A set $T$ of reals is bounded iff $T\subset [-r,r]$ for some $r>0.$ A sequence $(x_n)_{n\in N}$ is bounded iff $\{x_n:n\in N\}\subset [-r,r]$ for some $r>0.$

A convergent sequence is necessarily bounded because, if $\lim_{n\to \infty}x_n= l$ then the set $\{n\in N: |l-x_n|\geq 1\}$ is finite, so $\{x_n:n\in N\}$ is a subset of the bounded set $(-1+l,1+l)\cup \{x_n: n\in S\}.$

A non-convergent sequence may still be bounded,e.g. if $x_{2 n}=0$ and $x_{2 n-1}=1$ for each $n\in N.$ But if a sequence is unbounded it cannot be convergent.

For $x>1,$ let $x=1+y$ with $y>0.$ There are several ways to show that $(1+y)^n\geq 1+y n$ for any $y>0$ and any $n\in N.$ From this we see that $(x^n)_{n\in N}$ is an unbounded sequence and therefore not convergent.

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Assume that $x > 1$. Furthermore, assume for the sake of contradiction that $\left\{x^n\right\}_{n=1}^\infty$ converges to a limit $L$. That is $$ \lim_{n\to\infty}x_n = L.$$

We know from rules of exponentiation of real numbers that $x^{n+1} > x^n$ since x > 1. This means that our sequence $\left\{x^n\right\}_{n=1}^\infty$ is increasing. This means that $x^n \leq L$ for all $n \geq 1$ (This is because, if for some $n$, $x^n > L$ then $x^{n+1} > L$, $x^{n+2} > L$ and so on, hence not converging to $L$).

Now, we use the fact that $x^{n+1} = x\times x^n$. Using limit laws, we get that $$ \lim_{n\to\infty} x^{n+1} = \left(\lim_{n\to\infty} x\right) \times \left(\lim_{n\to\infty} x^n\right).$$

So we get the equality $L = xL$. Rearranging yields $L(1 - x) = 0$. At this point, either $L = 0$ or $(1 - x) = 0$. If $L = 0$, then $L$ cannot be a limit point, since $L = 0 < 1 < x = x^1 < x^n$ for all $n > 1$. So, the only other option is that $(1-x) = 0$, so $x = 1$. But this contradicts the fact that $x > 1$. Hence, our assumption must be wrong: so $\left\{x^n\right\}_{n=1}^\infty$ does not converge to a real number $L$.

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