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I came across a question which basically said that we needed to use the Binomial theorem to show that if $x_{n} > 0$, and $(1+x_{n})^{n}$ = n, then $x_{n}^{2} \le \frac{2}{n}$. Conclude that $n^\frac{1}{n} \to1$.

I am so confused as to how to proceed with this since it seems that we apply the Binomial Theorem to $(1+x_{n})^{n}$ but since we don't have an exact value of n, how do I expand this? And even if I do, how to deduce from there than $x_{n}^{2} \le \frac{2}{n}$? Can someone please help? Thanks!

Arctic Char
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1 Answers1

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The binomial theorem says

$$ (1+x_n)^n = \sum_{k=0}^n \binom{n}{k} x_n^k $$

You are assuming that this is equal to $n$ (that is what you said in the question). So set this equal to $n$ and solve for the $x_n^2$ term on the right-hand side: $$ \binom{n}{2} x_n^2 = n - 1 - nx_n - \binom{n}{3}x_n^3 - \cdots $$ The point is that since all the terms on the right being subtracted are positive (because $x_n > 0$ by your assumption), the right hand side is less than $n-1$. Now use the formula for $\binom{n}{2}$ and manipulate this inequality:

$$ \begin {align*} \binom{n}{2} x_n^2 &\leq n-1 \\ \frac{n!}{2(n-2)!} x_n^2 &\leq n-1 \\ x_n^2 &\leq (n-1) \cdot \frac{2(n-2)!}{n!} \end {align*} $$ If you simplify the right-hand side, this is equal to $\frac{2}{n}$.

Nick
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  • I see, but how can we conclude from this that $n^{\frac{1}{n}}$ tends to 1? –  Sep 15 '20 at 22:35
  • $n^{1/n} = 1+x_n$ by your assumption, so take the limit of $(1+x_n)$, using the inequality you now know: $x_n^2 \leq 2/n$. – Nick Sep 15 '20 at 22:37
  • Sorry, I am a little confused. What do you mean? –  Sep 15 '20 at 23:19
  • For all $n > 2$, $n = (1+x_n)^n > \frac{n(n-1)}{2}x_n^2 \implies 0 < n^{1/n} -1 = x_n < \sqrt{\frac{2}{n-1}}$ is probably what is being shown here. – RRL Sep 15 '20 at 23:51
  • Yes, essentially what @RRL said. Use the squeeze theorem.. If $x_n^2 \leq 2/n$, then $n^{1/n} = 1 + x_n \leq 1 + \sqrt{2/n}$. So you have the inequality $1 \leq n^{1/n} \leq 1 + \sqrt{2/n}$. – Nick Sep 16 '20 at 16:36