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I am trying to evaluate $$\lim\limits_{x\to \infty}x^2 \left( a^{1 \over x} - a^{1 \over x+1} \right)$$ for $a>0$.

The idea: $$a^{1 \over x} = e^{\ln{a^{1 \over x}}}=e^{{1 \over x}\ln{a}} = 1 + {\ln{a} \over x} + {1 \over 2}{(\ln{a})^2 \over x^2}+{1 \over 6}{(\ln{a})^3 \over x^3}+\dots$$

$$a^{1 \over x}-a^{1 \over x+1} = e^{\ln{a^{1 \over x}}}-e^{\ln{a^{1 \over x+1}}}=e^{{1 \over x}\ln{a}}-e^{{1 \over x+1}\ln{a}} = \ln{a} \left({1\over x} - {1 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ {1\over x^2} - {1 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x^3} - {1 \over (x+1)^3}}\right)+\dots$$

$$x^2 \left(a^{1 \over x}-a^{1 \over x+1}\right) =\ln{a} \left(x - {x^2 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ 1 - {x^2 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x} - {x^2 \over (x+1)^3}}\right)+\dots$$

This is where I get stuck. I suppose the whole sum should converge to $\ln{a}$ but can't quite find the way to get there. I would appreciate any comments.

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    You seem to have some issue with your Taylor expansion of $\exp$. When $u\to 0$, we have $e^u = 1+{\color{red} u} +\frac{u^2}{2} + o(u^2)$. – 2017-01-16

2 Answers 2

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You seem to have some issue with your Taylor expansion of $\exp$. Here, we will only need it to second or third order, but you forgot the linear term: when $u\to 0$, we have $$e^u = 1+\boxed{\color{red} u} +\frac{u^2}{2} + o(u^2)$$ which is crucial here.

Using this, we get, since $\frac{\ln a}{x} \xrightarrow[x\to\infty]{}0$, $$\begin{align} a^{\frac{1}{x}} &= e^{\frac{\ln a}{x}} = 1+\frac{\ln a}{x}+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \tag{1}\\a^{\frac{1}{x+1}}&= e^{\frac{\ln a}{x+1}} = 1+\frac{\ln a}{x+1}+\frac{\ln^2 a}{(x+1)^2} + o\left(\frac{1}{x^2}\right) \\ &= 1+\frac{\ln a}{x}\left(1-\frac{1}{x} + o\left(\frac{1}{x}\right) \right)+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \\ &= 1+\frac{\ln a}{x}-\frac{\ln a}{x^2}+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \tag{2} \end{align}$$ using also that $\frac{1}{1+u} = 1-u+o(u)$ when $u\to 0$ (in the third line). Thus, the difference of the two will be $$ a^{\frac{1}{x}} - a^{\frac{1}{x+1}} = \frac{\ln a}{x^2} + o\left(\frac{1}{x^2}\right) $$ from (1) and (2) (the other terms cancel), and the limit will then be $$ \lim_{x\to\infty}x^2\left( a^{\frac{1}{x}} - a^{\frac{1}{x+1}}\right) = \lim_{x\to\infty}x^2\left( \frac{\ln a}{x^2} + o\left(\frac{1}{x^2}\right) \right) = \boxed{\ln a} $$ as desired.

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    Regarding the first line: at the time I wrote this answer (before the OP edited the question to address my comment), indeed the linear term in the expansion was missing. – 2017-01-16
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I got it with a different approach. Take the logarithm of the limit,

$\ln(x^2) + \ln(a^{1/x} - a^{1/(1+x)}) = \ln(x^2) + \frac{\ln(a)}{x} + \ln(1 - a^{1/(1+x)-1/x})$

$\ln(x^2) + \frac{\ln(a)}{x} + \ln(1-e^{\frac{-\ln(a)}{x(1+x)}}) $

Since $\frac{-\ln(a)}{x(1+x)} \rightarrow 0$ as $x \rightarrow \infty$ we take a Taylor expansion of $e^{\frac{-\ln(a)}{x(1+x)}}$.

$\ln(x^2) + \frac{\ln(a)}{x} + \ln(1-(1-\frac{\ln(a)}{x(1+x)} + \mathcal{O}(\frac{1}{x^4})) $

$\ln(x^2) + \frac{\ln(a)}{x} + \ln(\frac{\ln(a)}{x(1+x)}) + \mathcal{O}(\frac{1}{x^2}) $

$\ln(\frac{x^2}{x(x+1)}) + \frac{\ln(a)}{x} + \ln(\ln(a)) + \mathcal{O}(\frac{1}{x^2}) \rightarrow \ln(\ln(a))$ as $x \rightarrow \infty$

Thus the limit itself is $\ln(a)$.