Series expansion of $\ln \left( 1 + \frac{1}{n} \right)^n $

Question

I'm working this problem out of an old algebra book, so part of the challenge is not to use Taylor series or anything from calculus (beyond the most basic notion of limits).

Apparently the solution is,

$$\ln \left(1 + \frac{1}{n} \right)^n = 1 - \frac{1}{2(n + 1)} - \frac{1}{(2)(3)(n + 1)^2} - \dots $$

My question is this: If

$$ \ln \left( 1 + x \right) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$$

why can't I simply substitute $\frac{1}{n}$ for $x$ to get,

$$ \ln \left( 1 + \frac{1}{n} \right)^n = n\ln \left(1 + \frac{1}{n} \right) = n \left( \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \dots \right) = 1 - \frac{1}{2n} + \frac{1}{3n^2} \dots $$

?

Answer 1

You almost figured it out in the comments. Indeed, \begin{align*} \ln \!\left( {1 + \frac{1}{n}} \right)^n & = - (n + 1)\ln \left( {1 - \frac{1}{{n + 1}}} \right) + \ln \left( {1 - \frac{1}{{n + 1}}} \right) \\ & = (n + 1)\sum\limits_{k = 1}^\infty {\frac{1}{k}\frac{1}{{(n + 1)^k }}} - \sum\limits_{k = 1}^\infty {\frac{1}{k}\frac{1}{{(n + 1)^k }}} \\ & = \sum\limits_{k = 1}^\infty {\frac{1}{k}\frac{1}{{(n + 1)^{k - 1} }}} - \sum\limits_{k = 1}^\infty {\frac{1}{k}\frac{1}{{(n + 1)^k }}} \\ & = \sum\limits_{k = 0}^\infty {\frac{1}{{k + 1}}\frac{1}{{(n + 1)^k }}} - \sum\limits_{k = 1}^\infty {\frac{1}{k}\frac{1}{{(n + 1)^k }}} \\ & = 1 + \sum\limits_{k = 1}^\infty {\frac{1}{{k + 1}}\frac{1}{{(n + 1)^k }}} - \sum\limits_{k = 1}^\infty {\frac{1}{k}\frac{1}{{(n + 1)^k }}} \\ & = 1 + \sum\limits_{k = 1}^\infty {\left( {\frac{1}{{k + 1}} - \frac{1}{k}} \right)\frac{1}{{(n + 1)^k }}} \\ & = 1 - \sum\limits_{k = 1}^\infty {\frac{1}{{k(k + 1)}}\frac{1}{{(n + 1)^k }}} , \end{align*} as desired. I used the Maclaurin series of $\ln(1+x)$ (wich converges for $-1<x<1$) with $x=-\frac{1}{n+1}$ though.

Answer 2

The way I like to derive the series for $\ln(1+x)$ is this:

$\dfrac{1-x^{n+1}}{1-x} =\sum_{k=0}^n x^k $ so $\dfrac{1}{1-x} =\sum_{k=0}^n x^k+\dfrac{x^{n+1}}{1-x} $ so $\dfrac{1}{1+x} =\sum_{k=0}^n (-x)^k+\dfrac{(-x)^{n+1}}{1+x} $ or $\dfrac{1}{1+x} =\sum_{k=0}^n (-1)^kx^k+\dfrac{(-1)^{n+1}x^{n+1}}{1+x} $.

Integrating from $0$ to $t$,

$\begin{array}\\ \ln(1+t) &=\int_0^t\dfrac{dx}{1+x}\\ &=\sum_{k=0}^n (-1)^k\int_0^tx^kdx+(-1)^{n+1}\int_0^t\dfrac{x^{n+1}}{1+x}dx\\ &=\sum_{k=0}^n (-1)^k\dfrac{t^{k+1}}{k+1}+(-1)^{n+1}\int_0^t\dfrac{x^{n+1}}{1+x}dx\\ &=\sum_{k=1}^{n+1} (-1)^{k-1}\dfrac{t^{k}}{k}+(-1)^{n+1}\int_0^t\dfrac{x^{n+1}}{1+x}dx\\ \end{array} $

so, replacing $n+1$ by $n$,

$\begin{array}\\ \ln(1+t)-\sum_{k=1}^{n} (-1)^{k-1}\dfrac{t^{k}}{k} &=(-1)^{n}\int_0^t\dfrac{x^{n}}{1+x}dx\\ &=(-1)^{n}E_n(t)\\ \end{array} $

where $E_n(t) =\int_0^t\dfrac{x^{n}}{1+x}dx $.

If $t > 0$,

$\begin{array}\\ E_n(t) &=\int_0^t\dfrac{x^{n}}{1+x}dx\\ &<\int_0^tx^{n}dx\\ &=\dfrac{t^{n+1}}{n+1}\\ \text{and}\\ E_n(t) &=\int_0^t\dfrac{x^{n}}{1+x}dx\\ &>\int_0^t\dfrac{x^{n}}{1+t}dx\\ &=\dfrac{t^{n+1}}{(n+1)(1+t)}\\ \end{array} $

so, if $0 < t < 1$, $E_n(t) \to 0$ and the series converges.