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.)
 A: Using the Taylor expansion of $\log(1+x)$ we see that
$$
1-n\log\Bigl(\frac{n+1}{n}\Bigr)=1-n\Bigl(\frac{1}{n}-\frac{1}{2\,n^2}+O(n^{-3})\Bigr)=\frac{1}{2\,n}+a_n,\quad a_n=O(n^{-2}).
$$
Then
$$
\sum_{n=1}^\infty(-1)^{n+1}\Bigl(1-n\log\Bigl(\frac{n+1}{n}\Bigr)\Bigr)=\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{2\,n}+\sum_{n=1}^\infty(-1)^{n+1}a_n.
$$
The first series on the right hand side converges by Leibniz's test, and the second is absolutely convergent since $|(-1)^{n+1}a_n|\le C\,n^{-2}$ for some constant $C>0$.
A: Look at $x \mapsto 1-x\log\left(\frac{x+1}{x}\right)$. Its second derivative is positive so its first derivative is increasing. It first derivative tends to $-\infty$ in when $x$ goes to zero and to $0$ when $x$ goes to $\infty$, thus it is negative, and the initial function is decreasing. Consider its restriction to $\mathbb{N}$ then.
A: 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.
A: Recall that for all $x>0$
$$
1 - \frac1x \leq \log x \leq x-1 \tag{1}
$$
If $x=\frac{n+1}{n}$, then $x>1$ and $n=\frac{1}{x-1}$. Multiplying $({1})$ by  $-n=\frac{1}{1-x}<0$ and then adding $1$ we have 
$$
1-n+\frac{n}{x}\geq1-n\log x \geq 1-nx+n
$$
or
$$
1-n+\frac{n^2}{n+1}\geq1-n\log \frac{n+1}{n} \geq 1-n\frac{n+1}{n}+n=0
$$ 
A more explicit estimate would be
$$
1-\frac{1}{n+1}\geq1-n\log \frac{n+1}{n} \geq 0 \tag{2}
$$ 
By the criterion of convergence of alternating series we have that the series 
$
\sum_{n = 1}^{\infty} 
(-1)^{n + 1} 
\left( 1 -  \frac{1}{n+1}\right)
$
converges. By the criterion of comparison for alternating series, it follows from $(2)$ that the series
$$ 
\sum_{n = 1}^{\infty} (-1)^{n+1}
\left( 1 - n \log \left( \frac{n + 1}{n} \right) \right)
$$
converges.
