<s>[INST] Determining and (dis)proving if $ \sum_{n = 1}^{\infty} (-1)^{n + 1} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) $ converges I am trying to determine if $ \sum_{n = 1}^{\infty} (-1)^{n + 1} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) $ converges using an alternating series test. The test in question requires me to prove $ 1 - n \log \left( \frac{n + 1}{n} \right) $ is decreasing and that $ \lim_{n \to \infty} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) = 0 $ to prove this series is convergent. 
My intutition says that this series is convergent because $ n \log \left( \frac{n + 1}{n} \right) $ will tend towards 1 as n goes to infinity (due to the definition of e). I am having trouble proving the sequence $ 1 - n \log \left( \frac{n + 1}{n} \right) $ is decreasing.
I set up the inequality $ 1 - (n + 1) \log \left( \frac{n + 2}{n + 1} \right) \leq 1 - n \log \left( \frac{n + 1}{n} \right) $ but I feel like I am stuck on some simple algebra. Any hints for proving the sequence is nonincreasing? Or am I just wrong?
EDIT: Could this be done using absolute convergence?
EDIT2: I am seeing some really great answers, but I am trying to prove this without calculus. (No derivatives or Taylor series expansions.)
 [/INST] Notice that $$\lim_{n\to\infty} n \ln\left( \frac{n + 1}{n} \right)=\lim_{n\to\infty}  \ln\left( 1+ \frac{1}{n} \right)^n=\ln e=1.$$
So $\lim_{n \to \infty} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) = 0$ as required and we are left with monotonicity:
$$
1- (n + 1) \log \left( \frac{n + 2}{n + 1} \right) \leq 1- n \log \left( \frac{n + 1}{n} \right)\\
  \log \left( \frac{n + 2}{n + 1} \right)^{n+1} \geq   \log \left( \frac{n + 1}{n} \right)^n\\
  \left( \frac{n + 2}{n + 1} \right)^{n+1} \geq   \left( \frac{n + 1}{n} \right)^n\\
 \left( 1+ \frac1{n+1} \right)^{n+1} \Biggr/ \left(1+ \frac1n \right)^n  \geq 1\\  
 \left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)^n \geq \frac{n+1}{n+2}\\ 
$$
Now use $ \left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)^n \geq \left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)=\frac{n(n+2)}{(n+1)^2}$ to conclude:
$$
\frac{n(n+2)}{(n+1)^2} \geq \frac{n+1}{n+2}\\
\frac{n(n+2)^2}{(n+1)^3}=\frac{n^3+4n^2+4n}{n^3+3n^2+3n+1} \geq 1\; ,\\
$$
for $n\geq 1$, which is not elegant, I know.
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