Find the limit

tobeno_1

let $a>0$ , then find the limit

$\lim_{n\to\infty}{n^2(\sqrt[n]{a}-\sqrt[n+1]{a})$

**Posted:** Sun Nov 01, 2009 8:43 pm

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Kent Merryfield

$\sqrt[n]{a}=e^{\frac{\ln a}{n}}=1+\frac{\ln a}{n}-\frac{\ln^2a}{2n^2}+O(n^{-3})$

$\sqrt[n+1]{a}=e^{\frac{\ln a}{n+1}}=1+\frac{\ln a}{n+1}-\frac{\ln^2a}{2(n+1)^2}+O(n^{-3})$

$\sqrt[n]{a}-\sqrt[n+1]{a}=\ln a\left(\frac1n-\frac1{n+1}\right)-\frac{\ln^2a}{2}\left(\frac1{n^2}-\frac1{(n+1)^2}\right)+O(n^...$

$\sqrt[n]{a}-\sqrt[n+1]{a}=\frac{\ln a}{n(n+1)}+O(n^{-3})$

$\lim_{n\to\infty}n^2\left(\sqrt[n]{a}-\sqrt[n+1]{a}\right)=\ln a.$

**Posted:** Sun Nov 01, 2009 9:46 pm

*210* · Hodge Conjecture Offline Joined: 07 Jul 2008 Posts: 53

Hello

We can also use " théorème des accroissements finis " for the function $f(x)=a^{\frac{1}{x}}$ in the intervalle $[n,n+1]$

We get the same result

**Posted:** Wed Nov 04, 2009 1:58 am

Kent Merryfield

In English, that would be the "mean value theorem."

If $f(x)=a^{\frac1x}$ then $f'(x)=-\frac{a^{\frac1x}\ln a}{x^2}$

So $f(n)-f(n+1)=-1\cdot f'(\xi)$ for some $n<\xi<n+1.$

$n^2(\sqrt[n]{a}-\sqrt[n+1]{a})=n^2\cdot\frac{a^{\frac1{\xi}}\ln a}{\xi^2}$

as $n\to\infty,$ $a^{\frac1{\xi}}\to 1$ and $\frac{n^2}{\xi^2}\to 1.$

**Posted:** Wed Nov 04, 2009 8:07 am

bgbgbgbg · Hodge Conjecture Offline Joined: 17 Sep 2009 Posts: 95

the limit is
lna

**Posted:** Wed Nov 04, 2009 9:14 pm