The conclusion I need to argue for is:
$ \forall \varepsilon \gt 0 \,\exists N \in \mathbb{N};\ \forall n \gt N \implies \vert \sqrt[n]{n}-1 \vert \lt \varepsilon $
In the book "Mathematical analysis and it's inherent nature" the author defined a sequence $a_{n}$ such that $a_{n} = \sqrt[n] n -1$ and then tries to prove that $\lim_{n\rightarrow \infty}a_{n}=0.$
to show that it uses the binomial theorem;$$
n=(1 + a_{n})^{n}=1+na_n + \frac{1}{2}n(n-1)a_{n}^2+...+a_{n}^n \ge 1+na_{n}+\frac{1}{2}n(n-1)a_{n}^2
$$my question is how to calculate the so called "simple calculation" to show $\ a_{n}^2 \le \frac{2}{n}$.
my effort on deriving the wanted inequality is:
$$ \begin{align}
1+na_{n}+\frac{1}{2}&n(n-1)a_{n}^2 \ \le \ n \\
a_n ^2 \le \frac{2(n-1-na_n )}{n(n-1)} &\le \frac{2(1-a_n)}{(n-1)} \le \frac{2}{n-1}
\end{align}$$
but it seems like I can't go further.
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1You can forget about $na_n$ which gives $n \ge 1 + \frac{n(n-1)}{2} a_n^2$ – Presage Dec 14 '23 at 14:14
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1Split the numerator into $2(n-1) - 2na_n$. – SuhailSherif Dec 14 '23 at 14:22
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This has been asked and answered many times: https://math.stackexchange.com/questions/linked/115822 – found with Approach0 – Martin R Dec 14 '23 at 19:13
2 Answers
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Too messy. You can do it with only:
$$n\ge\frac12n(n-1)a_n^2\implies a_n\le\sqrt{\frac2{n-1}}$$
and since the rightmost term clearly tends to zero as $\;n\to\infty\;$ we're done...even with the squeeze theorem, if you want.
DonAntonio
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As $a_n \geq 0$, we get \begin{align*} a_n^2 \leq \frac{2(n-1-na_n)}{n(n-1)} \leq \frac{2(n-1)}{n(n-1)} = \frac{2}{n}. \end{align*}
Tobi
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