A more difficult problem which can be solved by the help of § 69 arises when \(\phi(n) = \{1 + 1/n\}^{n}\).

It follows from the binomial theorem1 that \[\begin{gathered} \begin{aligned} \biggl(1 + \frac{1}{n}\biggr)^{n} &= 1 + n \cdot \frac{1}{n} + \frac{n(n – 1)}{1\cdot2}\, \frac{1}{n^{2}} + \dots + \frac{n(n – 1)\dots (n – n + 1)}{1\cdot2\dots n}\, \frac{1}{n^{n}}\\ &= 1 + 1 + \frac{1}{1\cdot2} \biggl(1 – \frac{1}{n}\biggr) + \frac{1}{1\cdot2\cdot3} \biggl(1 – \frac{1}{n}\biggr) \biggl(1- \frac{2}{n}\biggr) + \dots\\ \end{aligned} \\ + \frac{1}{1\cdot2\dots n} \biggl(1 – \frac{1}{n}\biggr) \biggl(1 – \frac{2}{n}\biggr)\dots \biggl(1 – \frac{n – 1}{n}\biggr).\end{gathered}\]

The \((p + 1)\)th term in this expression, viz. \[\frac{1}{1\cdot2\dots p} \left(1 – \frac{1}{n}\right) \left(1 – \frac{2}{n}\right)\dots \left(1 – \frac{p – 1}{n}\right),\] is positive and an increasing function of \(n\), and the number of terms also increases with \(n\). Hence \(\left(1 + \dfrac{1}{n}\right)^{n}\) increases with \(n\), and so tends to a limit or to \(+\infty\), as \(n \to \infty\).

But \[\begin{aligned} \left(1 + \frac{1}{n}\right)^{n} &< 1 + 1 + \frac{1}{1\cdot2} + \frac{1}{1\cdot2\cdot3} + \dots + \frac{1}{1\cdot2\cdot3 \dots n}\\ &< 1 + 1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n-1}} < 3.\end{aligned}\]

Thus \(\left(1 + \dfrac{1}{n}\right)^{n}\) cannot tend to \(+\infty\), and so \[\lim_{n \to\infty} \left(1 + \frac{1}{n}\right)^{n} = e,\] where \(e\) is a number such that \(2 < e \leq 3\).


  1. The binomial theorem for a positive integral exponent, which is what is used here, is a theorem of elementary algebra. The other cases of the theorem belong to the theory of infinite series, and will be considered later.↩︎

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