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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.

Asaf Karagila
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mboratko
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  • Are you sure that Apostol doesn't discuss the rate of convergence of $(1+(1/n))^n$ earlier in the book? I know the problems in Apostol are hard, but I'm surprised that he would spring this one on people without having paved the way first. – Gerry Myerson Oct 09 '11 at 23:53
  • 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. – mboratko Oct 10 '11 at 00:01

4 Answers4

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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?

davidlowryduda
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  • What is the generalization? I searched, but it relayed me to the product rule for differentiation. – mboratko Oct 10 '11 at 00:10
  • Where does your original inequality come from? Is that something used commonly, I haven't seen it before. – mboratko Oct 10 '11 at 05:40
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    @process: It's just a factoring trick. The numerator factors, and since each a is less than b and there are n + 1 terms, we get that inequality very quickly. Does that make sense? – davidlowryduda Oct 10 '11 at 06:11
  • Yes, it does. Thanks! – mboratko Oct 10 '11 at 06:16
  • @davidlowryduda, how did you recognise this trick, if its not a secret? I thought about it for a while - I just do not see a way to connect this identity, which I knew, to the $(1 + 1/n)$..I mean if one does not know that this helps, then it is impossible to apply it... – John Dec 05 '19 at 23:30
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It's easy to show that $(1+\frac{x}{n})^n$ is increasing for all $x>0$:

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

kahen
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  • OK, I get this. Any suggestions for handling absolute convergence/divergence? – mboratko Oct 09 '11 at 23:58
  • @kahen, Wait, in a line before the last one, the term with n+1 power is actually from 0 to 1. Thus, the last line is not necessarily smaller than the line before it! I do not think this inequality can be formed. – John Dec 05 '19 at 23:21
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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.

marty cohen
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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. – mboratko Oct 10 '11 at 01:14
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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.

mboratko
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