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If $\vert x \vert < 1$ then I want to show $\lim_{n\rightarrow \infty} (n+1)\cdot x^{n\cdot n!} < 1$

It makes perfectly sense in my world, because the factor $x^{n\cdot n!}$ is smaller than the factor $(n+1)$ when n goes to infinity. I have tried to use L'Hoptial but it doesn't work. Then I tried to find an example of an expression which is greater than $(n+1)\cdot x^{n\cdot n!}$ but still smaller than 1, when n goes to infinity. But all the examples I have found diverges. Now I'm stuck - any ideas?

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Use the squeeze theorem by noting $r^{(n+1)}\leq {r^{(n+1)!}} $ with $r=1/x$. –  Pedro Tamaroff May 26 '12 at 16:27

2 Answers 2

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Take $\,a\in\mathbb{R}\,,\,|a|<1\,$ , and use L'Hospital to prove that $$\lim_{x\to\infty}(x+1)a^x=\lim_{x\to\infty}\frac{x+1}{\frac{1}{a^x}}=0$$Now just observe that $\,(n+1)a^{nn!}\leq (n+1)a^n$ ...

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Ok. Simple! My only problem with this is the L'hoptial (I suppose you mean hoptial and not hospital :) ) part. If you use L'hopital then you end up with: $\lim_{x\rightarrow \infty} \frac{1}{-\ln (a) \cdot a^{x}}$ a could be a negative number? and ln to a negative number makes no sense? –  characters May 26 '12 at 16:40
    
Indeed, no sense within the real numbers, which is what we need, but you can limit yourself to positive $a$ and then generalize to $\,-1<a<0\,$ by simply multiplying the whole expression by -1 (since, as it happens, the limit is zero) –  DonAntonio May 26 '12 at 16:49

The case ${x}=0$ is clear. For the rest of the values we use the fact that $\vert x \vert < 1$, and we may replace ${x}$ by $\frac{1}{t}$ where $\vert t \vert > 1$. Lastly, we get that: $$\lim_{n\rightarrow \infty} \frac{n+1}{t^{{n} \cdot {n!}}} =\frac{polynomial \space growth \space behaviour}{exponential \space growth \space behaviour} =0$$

The proof is complete.

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