# Leibniz standard for rotating collection

Given the collection : $\sum_{n=0}^{\infty}(-1)^{n}(\sqrt[n]{n} - 1)^{n}$. Does the collection merge?

Try to remedy (could be wrong) :

$(\sqrt[n]{n} - 1)^{n}> (1+\frac{1}{n})^{n}$

$(1+\frac{1}{n})^{n} \to e \Rightarrow (\sqrt[n]{n} - 1)^{n}$ lower - bounded by $e$. Based Upon Leibniz Criterion the series $\{A_n\}$ (in our instance, $(\sqrt[n]{n} - 1)^{n}$) is monotone lowering, yet its restriction is not $0$ at boundless $\Rightarrow$ collection deviate.

Is it sufficient to claim that given that the series is lower - bounded, the restriction of it at infinite is not $0$, or should I in fact compute the restriction of the series?

Write your collection as $\sum_{n = 1}^\infty (-1)^n a_n$ where $a_n = (n^{1/n} - 1)^n$. Keep in mind that $(a_n)$ is monotonically lowering to absolutely no and also all terms declare, therefore by the rotating collection examination (or Leibniz' examination) the collection merges.

Your lower bound for $(n^{1/n} -1)^n$ is not deal with.

Tip : To show that $(n^{1/n} -1)^n \rightarrow 0$ write $n^{1/n} = 1 + r_n$ and also make use of the binomial theory to show that for $n \ge 2$ we have $r_n < \sqrt{2/(n-1)}.$

Hint2 : When you expand $(1+r_n)^n$ the vital term in the development is the one in ${r_n}^2 .$

In this means you can show that the collection merges definitely, you do not actually require the Leibniz standard although they will certainly get the job done.

The alternatingness of the collection is something of a false trail as the collection merges definitely. By the origin examination, to show this it is adequate to show that $\lim_{n \rightarrow \infty} |n^{1 \over n} - 1| = 0$. To put it simply, it is adequate to show that $\lim_{n \rightarrow \infty} n^{1 \over n} = 1$.

There are a couple of means to show this restriction remains in reality $1$. One means is to keep in mind that $\ln (n^{1 \over n}) = {\ln(n) \over n}$, and also the last is attended most likely to absolutely no as $n$ mosts likely to infinity making use of L'Hopital is regulation. Given that the all-natural log of the $n$th regard to the series mosts likely to absolutely no, the $n$th regard to the series mosts likely to $e^0 = 1$.

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