I have the next limit: $$\lim\limits_{n\to \infty }\left(\sqrt[n]{\left(\frac{n!-1}{n!+1}\right)^{\left(n+1\right)!}}\right)$$
I had done some steps and simplified it to:
$$\lim\limits_{n\to \infty }\left(1-\frac{2}{n!+1}\right)^{(n+1)(n-1)!}=\\
\lim\limits_{n\to \infty }\left(1-\frac{1}{(n(n-1)!+1)\cdot 0.5}\right)^{(n+1)(n-1)!}$$
And my final result is:
$$\lim\limits_{n\to \infty }\left(\frac{1}{e}\right)^{\frac{3n+2+\frac{1}{(n-1)!}}{2}}$$
My question is what happens to $\frac{1}{(\infty -1)!}$? Is it 0?
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