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Determine the convergence of the series (-1)n (e – (1+1/n)n)

Consider the series

    \[ \sum_{n=1}^{\infty} (-1)^n \left( e - \left(1 + \frac{1}{n} \right)^n \right). \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.


The series converges conditionally.

Proof. First, we use the Leibniz rule to show that the series converges. We have

    \[ \lim_{n \to \infty} \left( e - \left( 1 + \frac{1}{n} \right)^n \right) = 0 \]

so the terms are going to 0. Then, we need to show that the terms are monotonically decreasing. We will use the Arithmetic Mean-Geometric Mean inequality to do this (which we proved in this exercise, Section I.4.10, Exercise #20). This inequality states that for positive real numbers x_1, \ldots, x_n we have

    \[ (x_1 x_2 \cdots x_n)^{\frac{1}{n}} \leq \frac{x_1 + x_2 + \cdots + x_n}{n} \]

(taking p = 1 in the exercise) with equality only if x_1 = x_2 = \cdots = x_n. So, for the problem at hand we consider the n+1 positive real numbers

    \[ x_1 = 1, \qquad x_2 = x_3 = \cdots = x_{n+1} = 1 + \frac{1}{n}. \]

Then these are not all equal so we have strict inequality in the AM-GM inequality, giving us

    \begin{align*}   && \left( x_1 x_2 \cdots x_{n+1} \right)^{\frac{1}{n+1}} &< \frac{x_1 + x_2 + \cdots + x_{n+1}}{n+1} \\[9pt]  \implies && \left( 1 + \frac{1}{n} \right)^{\frac{n}{n+1}} &< \frac{1 + n \left( 1 + \frac{1}{n} \right)}{n+1} \\[9pt]  \implies && \left( 1 + \frac{1}{n} \right)^{\frac{n}{n+1}} &< 1 + \frac{1}{n+1} \\[9pt]  \implies && \left( 1 + \frac{1}{n} \right)^n &< \left( 1 + \frac{1}{n+1} \right)^{n+1}. \end{align*}

Thus, the n+1st term is greater than the nth so the sequence whose terms are giving by \left( 1 + \frac{1}{n} \right)^n is strictly increasing. Therefore, the sequence with terms

    \[ e - \left( 1 + \frac{1}{n} \right)^n \]

is strictly decreasing. Hence, by the Leibniz rule the series

    \[ \sum_{n=1}^{\infty} (-1)^n \left( e - \left( 1 + \frac{1}{n} \right)^n \right) \]

converges.

This convergence is conditional since

    \[ \sum_{n=1}^{\infty} \left| (-1)^n \left( e - \left(1 + \frac{1}{n} \right)^n \right) \right| = \sum_{n=1}^{\infty} \left( e - \left( 1+ \frac{1}{n} \right)^n \right). \]

Then, we use the limit comparison test to compare this series with the series \sum \frac{1}{n}. First, we make the substitution x = \frac{1}{n} to obtain

    \begin{align*}  \lim_{n \to \infty} \frac{e - \left( 1 + \frac{1}{n} \right)^n}{\frac{1}{n}} &= \lim_{x \to 0^+} \frac{e - (1+x)^{\frac{1}{x}}}{x} \\[9pt]  &= \lim_{x \to 0^+} \frac{-(1+x)^{\frac{1}{x}} \left( \frac{1}{x(1+x)} - \frac{1}{x^2} \log (1+x)}{1} \\[9pt]  &= \lim_{x \to 0^+} \frac{-(1+x)^{\frac{1}{x}} \left( \frac{x}{x+1} - \log (1+x) \right)}{x^2} \\[9pt]  &= \lim_{x \to 0^+} \left(-(1+x)^{\frac{1}{x}}\right)\left( \frac{\frac{x}{x+1} - \log(1+x)}{x^2} \right). \end{align*}

Now, the limit of the first term in the product is

    \[ \lim_{x \to 0^+} -(1+x)^{\frac{1}{x}} = -e. \]

For the second term in the product we have

    \begin{align*}  \lim_{x \to 0^+} \frac{\frac{x}{x+1} - \log(1+x)}{x^2} &= \lim_{x \to 0^+} \frac{\frac{1}{(x+1)^2} - \frac{1}{x+1}}{2x} \\[9pt]  &= \lim_{x \to 0^+} \frac{-x}{2x(1+x)^2} \\[9pt]  &= -\frac{1}{2}. \end{align*}

Therefore,

    \[ \lim_{n \to \infty} \frac{e - \left( 1 + \frac{1}{n} \right)^n}{\frac{1}{n}} = \frac{e}{2} > 1. \]

Hence, by the ratio test, the series diverges. So, the convergence of the series in the exercise is conditional. \qquad \blacksquare

2 comments

  1. Yunus Hermann says:

    Hi, I am extremly interested how one may show, that the, on the first glance quite similar series of 1/n*(e-(1+1/n)^n) converges. I think the best criterium to use might be the one derived from cauchy-schwarzes inequality. using this what would be left to show is, that the series of (e-(1+1/n)^n)^2 convergrs, which i think it does. i know this questionen is only loosly connected to the problem at hand but i am really desperate to find out!
    best regards

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