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.1.4 Title 15. Title IV, Part A, Sections 4007 to 4010, is effective as a statutory prohibition.Q:
Proving: $\lim_{n\to \infty} (1- \frac1n)^n=0$
I want to show that: $$\lim_{n\to \infty} (1- \frac1n)^n=0$$
I started to solve it this way:
I see the variable $x$ can be replaced by $\frac1n$ but when I try to:
I am stuck on what to do after $ln(1-x)$… I know that $ln$ is a logarithm, a function with many properties, but it is not a quotient, so:
What should I do next to reach the answer?
EDIT: I wanted to show that $\lim_{n\to \infty} (1- \frac1n)^n=0$
A:
Hint
Note that
$$
\left( 1- \frac 1n \right)^n = \left( 1 – \frac 1n \right)^{\frac{n(n-1)}{2}}
$$
So
$$
\left( 1 – \frac 1n \right)^n = e^{ -1} \to 0
$$
A:
First we know that: $\sum_{n=1}^\infty\frac1{n^k}=\frac1{k+1}\to0$ as $k\to\infty$, by comparison test. Now we know: $\frac{1}{n^k}=n^{\left(-k\right)}$ and $\left(1-n^{ -k}\right)^n=e^{n\log\left(1-n^{ -k}\right)}$. So we need to show that:
$\lim_{n\to\infty}e^{n\log\left(1-n^{ -k}\right)}\to0$. And: $\log\left(1-n^{ -k}\right)=n^{ -k}-\frac{n^{ -k}}{2}+…$ So:
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