I have a sequence $x_n$ and I want to prove that $\sqrt[n]{x_n}\le\sqrt[n+1]{x_{n+1}}$ for every $n$. The problem is I don't know how to handle the transition from $n$ to $n+1$ in the exponent. Are there nice relations/estimations/bounds of $\sqrt[n+1]{x}$ given $\sqrt[n]{x}$? This seems like a logical place to start.
An estimation of $\sqrt[n+1]{x}$ given $\sqrt[n]{x}$?
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calculus
inequality
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8It seems to me that the most natural estimation is $$(\sqrt[n]{x})^{n/(n+1)}.$$ All other bounds will depend on $x$ as well - I mean you cannot write $|\sqrt[n]{x} - \sqrt[n+1]{x}|\leq \varepsilon$ or $\frac{\sqrt[n]{x}}{\sqrt[n+1]{x}}\leq k$ for all $x$. – 2011-06-20
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0Your estimation seems one-sided... Am I missing something? – 2011-06-20
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1The sequence is presumably non-negative. One approach would be to see whether $x_{n}^{n+1} \le x_{n+1}^{n}$, avoiding roots. – 2011-06-20
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4Some *general* strategies have been mentioned, but there are many other possibilities. Could we be told the *specific* problem? The strategy for that one might work for similar ones. – 2011-06-20
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0Since you "have" a sequence $(x_n)_{n\geq0}$ it would help if you could give us some additional information about this sequence. Maybe it is then possible to crank out additional information about the relation between $x_n$ and $x_{n+1}$. – 2011-07-24