If in the first inequality (3) of § 74 we put \(r = 1/(n – 1)\), \(s = 1/n\), we see that \[(n – 1)(\sqrt[n-1]{\alpha} – 1) > n(\sqrt[n]{\alpha} – 1)\] when \(\alpha > 1\). Thus if \(\phi(n) = n(\sqrt[n]{\alpha} – 1)\) then \(\phi(n)\) decreases steadily as \(n\) increases. Also \(\phi(n)\) is always positive. Hence \(\phi(n)\) tends to a limit \(l\) as \(n \to \infty\), and \(l \geq 0\).

Again if, in the first inequality (7) of § 74, we put \(s = 1/n\), we obtain \[n(\sqrt[n]{\alpha} – 1) > \sqrt[n]{\alpha}\left(1 – \frac{1}{\alpha}\right) > 1 – \frac{1}{\alpha}.\] Thus \(l \geq 1 – (1/\alpha) > 0\). Hence, if \(\alpha > 1\), we have \[\lim_{n \to \infty} n(\sqrt[n]{\alpha} – 1) = f(\alpha),\] where \(f(\alpha) > 0\).

Next suppose \(\beta < 1\), and let \(\beta = 1/\alpha\); then \(n(\sqrt[n]{\beta} – 1) = -n(\sqrt{\alpha} – 1)/\sqrt[n]{\alpha}\). Now \(n(\sqrt[n]{\alpha} – 1) \to f(\alpha)\), and (Exs. XXVII. 10) \[\sqrt[n]{\alpha} \to 1.\] Hence, if \(\beta = 1/\alpha < 1\), we have \[n(\sqrt[n]{\beta} – 1) \to -f(\alpha).\] Finally, if \(x = 1\), then \(n(\sqrt[n]{x} – 1) = 0\) for all values of \(n\).

Thus we arrive at the result: the limit \[\lim n(\sqrt[n]{x} – 1)\] defines a function of \(x\) for all positive values of \(x\). This function \(f(x)\) possesses the properties \[f(1/x) = -f(x),\quad f(1) = 0,\] and is positive or negative according as \(x > 1\) or \(x < 1\). Later on we shall be able to identify this function with the Napierian logarithm of \(x\).

Example. Prove that \(f(xy) = f(x) + f(y)\). [Use the equations \[f(xy) = \lim n({\sqrt[n]{xy}} – 1) = \lim \{n(\sqrt[n]{x} – 1)\sqrt[n]{y} + n(\sqrt[n]{y} – 1)\}.]\]


$\leftarrow$ 74. Some algebraical lemmas Main Page 76-77. Infinite series $\rightarrow$
Close Menu