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?

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2019-12-02 02:49:06
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Answers: 3

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.

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2019-12-03 04:24:51
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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.

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2019-12-03 04:24:47
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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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2019-12-03 04:24:19
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