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My homework asks me to calculate (if it exists) the following limit:


My thinking is: $(-1)^n$ would, as we all know, oscillate between 1 and -1, meaning that $(1+(-1)^n)$ would be either $0$ or $2$. Thus, for all odd cases: $$\lim_{n\to\infty}{\frac{0^n}{n}}=0$$ And then, for all even cases: $$\lim_{n\to\infty}{\frac{2^n}{n}}$$ Using Cauchy: $$\lim_{n\to\infty}{^n\sqrt{\frac{2^n}{n}}}$$ $$\lim_{n\to\infty}{\frac{^n\sqrt{2^n}}{^n\sqrt{n}}}$$ $$\lim_{n\to\infty}{\frac{2}{1}} = 2$$

And then, it follows that $$\lim_{n\to\infty}{\frac{2^n}{n}} = \infty$$ Which means that our original expression... has no limit?

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up vote 1 down vote accepted

You are right. The sequence does not have a limit.

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Set $a_n=\frac{(1+(-1)^n)^n}{n}$

Observe that

$\limsup_{n\rightarrow\infty}a_n:=\lim_{n\rightarrow\infty}(\sup_{m\geq n}a_m)=\infty$


$\liminf_{n\rightarrow\infty}a_n:=\lim_{n\rightarrow\infty}(\inf_{m\geq n}a_m)=0$.

Since $\limsup_{n\rightarrow\infty}a_n \neq\liminf_{n\rightarrow\infty}a_n$, the limit of the sequence does not exist.

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