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How can I prove this inequality $$\sqrt[n]{x^n+\sqrt[n]{(2x)^n+\sqrt[n]{(3x)^n+\cdots}}}< \left(x+\frac{1}{n-1}\right)$$ if $n$ and $x$ are positive integer number $$x>=1$$ $$n>1$$

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  • $\begingroup$ What are the conditions of $x$ and $n$? $\endgroup$ – Idonknow Nov 14 '14 at 17:19
  • $\begingroup$ I think something is missing on the left hand side, since buy putting $x=1$ we can estimate it from below by $(1+2)^{1/n}$. Hence, we would get $3<(1+ 1/(n-1))^{n}$, which tends to $e=2.718281828...$ $\endgroup$ – AD. Nov 14 '14 at 17:38
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As A.D. noted, the inequality is wrong. So what is the correct upper bound?

Let $$ u_{km} = \dfrac{1}{kx}\sqrt[n]{(k x)^n + \sqrt[n]{((k+1) x)^n + \sqrt[n]{\ldots + \sqrt[n]{(mx)^n}}}}$$ with $x \ge 0$, $m \ge k \ge 1$, $n \ge 1 $, so the left side of your inequality is $\lim_{m \to \infty} x u_{1m}$.

We have $$ u_{km} = \sqrt[n]{1 + \dfrac{k+1}{k^n x^{n-1}} u_{k+1,m}} \le 1 + \dfrac{k+1}{n k^n x^{n-1}} u_{k+1,m}$$ with $u_{mm} = 1$. We get $$ u_{1m} \le \sum_{j = 0}^{m-1} \dfrac{(j+1)!}{(n x^{n-1})^j (j!)^n} < \sum_{j = 0}^{\infty} \dfrac{(j+1)!}{(n x^{n-1})^j (j!)^n} $$ Since $n \ge 2$, $(j+1)!/(j!)^n \le (j+1)!/(j!)^2 = (j+1)/j!$ and $x^{n-1} \ge x$ so $$ u_{1m} < \sum_{j=0}^\infty \dfrac{j+1}{(n x)^j j!} = \left(1 + \dfrac{1}{n x}\right) e^{1/(n x)}$$ Thus a correct inequality is $$ \sqrt[n]{x^n + \sqrt[n]{(2 x)^n + \sqrt[n]{\ldots }}} < \left(x + \dfrac{1}{n}\right) e^{1/(nx)}$$

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  • $\begingroup$ please, tell me if you can catch any positive integer number for both $x$ and $n$ not satisfy the inequality $\endgroup$ – E.H.E Nov 14 '14 at 18:50
  • $\begingroup$ Try $x = 1$, $n = 6$. The left side is greater than $3^{1/6}$, which is greater than $1 + 1/5$. $\endgroup$ – Robert Israel Nov 14 '14 at 18:59
  • $\begingroup$ you are right. Your justification is ok, thanks $\endgroup$ – E.H.E Nov 14 '14 at 19:06
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Since $$ a^n-b^n = (a-b)\sum_{k=0}^{n-1}a^k b^{n-k-1} $$ it follows that: $$ a-b = \frac{a^n-b^n}{\sum_{k=0}^{n-1}a^k b^{n-k-1}}$$ hence: $$\sqrt[n]{x^n+a}-x = \frac{a}{\sum_{k=0}^{n-1}(x^n+a)^{\frac{k}{n}}x^{n-k-1}}$$ and we have that: $$f_n(x)=-x+\sqrt[n]{x^n+\sqrt[n]{(2x)^n+\ldots}}$$ is a decreasing function for $x\geq 1$. So we have only to prove that: $$f_n(1)=-1+\sqrt[n]{1+\sqrt[n]{2+\ldots}}\leq\frac{1}{n-1}\tag{1}$$ that is equivalent to proving that: $$g_n(1)=1+\sqrt[n]{2+\sqrt[n]{3+\ldots}}\leq\left(1+\frac{1}{n-1}\right)^n.\tag{2}$$

Since the sequence given by $a_n=\left(1+\frac{1}{n-1}\right)^n$ decreases towards $e$ and $g_n(1)$ is also a decreasing sequence, it is sufficient to check that $(2)$ holds for $n=2$ and for $n=3$ we have: $$g_n(1)=1+\sqrt[n]{2+\sqrt[n]{3+\ldots}}\leq\left(1+\frac{1}{n-1}\right)^n\leq e.$$ So we have just to compute two limits with a reasonable accuracy: $$g_2(1)=3.09033\ldots<4,\qquad g_3(1)=2.54498\ldots<e.$$

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