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I am having a lot of trouble with these series questions. Up until this point, I had relatively little trouble with all the questions in the book. These seem to require knowledge about approximations of functions and other external experience-based knowledge, which I just don't have yet.

Determine convergence or divergence of the given series. In the case of convergence, determine whether the series converges absolutely or conditionally.

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

It's easy to see that

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

however, in order to apply Leibniz's Rule and show conditional convergence I need to show that the sequence is monotonically decreasing. This doesn't seem doable with straight inequalities, so I tried taking the derivative, which just resulted in an uninterpretable mess. This doesn't even begin to address the question of absolute convergence/divergence.

There are 54 of these questions... I must be missing something really fundamental if they all take this long.

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    I've done literally every problem in the book so far. He's referenced the limit of $(1+1/n)^n$, but nothing that I can find relating the rate of convergence.2011-10-10

4 Answers 4

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With this one, I would apply Leibniz's Rule. To do that, I would show that $(1 + \frac{1}{n})^n$ is monotonically increasing. There are many ways to do this, and I will give you one (my favorite that I've seen).

Note that $\dfrac{b^{n+1} - a^{n+1}}{b-a} < (n+1)b^n$ when $b > a \geq 0$

This means that $b^n [ (n + 1)a - n b] < a^{n+1}$ (just rearrange).

Then set $b = 1 + \frac{1}{n}$ and $a = 1 + \frac{1}{n+1}$, and we get the desired inequality.

It may not be the case that you are missing anything at all, really. Your intuition to use Leibniz's rule is a very good one - note also that in this section, there is a generalization of Leibniz's rule that is super handy. Sometimes, one must do very witty things to get through a question - if everything were simple arithmetic, it really just wouldn't be worth studying or fun, you know?

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    Yes, it does. Thanks!2011-10-10
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It's easy to show that $(1+\frac{x}{n})^n$ is increasing for all $x>0$:

$\begin{align} \frac{(1+\frac{x}{n+1})^{n+1}}{(1+\frac{x}{n})^n} &= (1+\frac{x}{n})\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1} \\\\ &= (1+\frac{x}{n})\left(\frac{n(n+1)+nx}{(n+1)(n+x)}\right)^{n+1} \\\\ &= (1+\frac{x}{n})\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1} \\\\ &= (1+\frac{x}{n})\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1} \\\\ &> (1+\frac{x}{n})(1-\frac{x}{n+x}) = \frac{n+x}{n} \frac{n}{n+x} = 1. \end{align}$

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    OK, I get this. Any suggestions for handling absolute convergence/divergence?2011-10-09
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$\begin{align} (1+1/n)^n &= \exp(n ln(1+1/n)) \\ &= \exp(n(1/n - 1/2n^2 + O(1/n^3)) \\ &= \exp(1-1/2n+O(1/n^2) \\ &= e \exp(-1/2n+O(1/n^2)) \\ &= e(1-1/2n+O(1/n^2)) \\ &= e-e/2n+O(1/n^2) \end{align} $ so $e - (1+1/n)^n = e/2n + O(1/n^2).$

Substituting, $ \sum_{n=1}^\infty (-1)^n\left[e-\left(1+\frac 1 n \right)^n\right] = \sum_{n=1}^\infty (-1)^n \left(e/2n + O(1/n^2) \right). $

Since $\sum 1/n$ diverges, the series converges, but not absolutely.

The estimates can be made more precise and rigorous, but I am feeling lazy.

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    This answer may very well be correct, however the book has not touched big-O notation yet. While little-o notation was discussed, and can of course be used in place of this, bringing these estimates out of the exponent is something that, while I have seen it used to explain these questions before, I do not truly understand it because it has not been discussed in this book.2011-10-10
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I think I have an answer for absolute divergence:

$\lim_{n\to\infty}\frac{e-(1+\frac 1 n )^n}{\frac 1 n}=\lim_{x\to 0^+} \frac{e-(1+x)^\frac 1 x}{x}=-\lim_{x\to0^+}\frac{(1+x)^\frac 1 x\left[\frac x {1+x}-\log(1+x)\right]}{x^2}$

It now helps to break the limits up a bit.

$\lim_{x\to0^+}(1+x)^\frac 1 x=e$ So now we focus on $\lim_{x\to0^+} \frac{\frac x {1+x}-\log(1+x)}{x^2}=\lim_{x\to0^+} \frac{\frac 1 {(1+x)^2}-\frac 1 {1+x}}{2x}=\lim_{x\to0^+}\frac{-x}{2x(1+x)^2}$ $=\lim_{x\to0^+}\frac{-1}{2(1+x)^2}=\frac{-1}{2}$ Now putting it all back together, $\lim_{n\to\infty}\frac{e-(1+\frac 1 n )^n}{\frac 1 n}=\frac e 2$ So, by the limit comparison theorem, the series (absolutely) diverges.