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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 $r=1/x$.2012-05-26

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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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    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 by simply multiplying the whole expression by -1 (since, as it happens, the limit is zero)2012-05-26
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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.