Series - Apostol Calculus Vol I, Section 10.20 #24 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.
 A: 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?
A: It's easy to show that $(1+\frac{x}{n})^n$ is increasing for all $x>0$: 
$$\begin{align}
\frac{(1+\frac{x}{n+1})^{n+1}}{(1+\frac{x}{n})^n} 
  &= (1+\frac{x}{n})\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1} \\\\
  &= (1+\frac{x}{n})\left(\frac{n(n+1)+nx}{(n+1)(n+x)}\right)^{n+1} \\\\
  &= (1+\frac{x}{n})\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1} \\\\
  &= (1+\frac{x}{n})\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1} \\\\
  &> (1+\frac{x}{n})(1-\frac{x}{n+x}) = \frac{n+x}{n} \frac{n}{n+x} = 1.
\end{align}$$
A: $$\begin{align}
(1+1/n)^n 
&= \exp(n ln(1+1/n)) \\
&= \exp(n(1/n - 1/2n^2 + O(1/n^3)) \\
&= \exp(1-1/2n+O(1/n^2) \\
&= e \exp(-1/2n+O(1/n^2)) \\
&= e(1-1/2n+O(1/n^2)) \\
&= e-e/2n+O(1/n^2)
\end{align}
$$
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.
A: 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.
