Inequality with Euler's totient Using the first 100,000 values of $\varphi(n)$ it seems that the following is true.
Let $\mathcal A$ be a finite subset of $\mathbb{N}$, $\forall n\in \mathbb{N} \setminus \mathcal{A}$, $\displaystyle \frac{1}{\varphi(2n)} -  \frac{1}{\varphi(2n+1)} \geqslant \frac{1}{2n\ln (2n)} $.
Is this true? Is there a stronger lower bound?
P.S.: I looked at Handbook of Number Theory I by Mitrinović and Sándor which has a lot of info about $\varphi (n)$ but it doesn't appear there.
 A: As already mentioned in the comments, one cannot expect inequalities of the form $$\varphi(an+b)>\varphi(cn+d)$$
(provided $ad\neq bc$) to hold for all n. In fact, as is proved here for any polynomials $\{a_ix+b_i\}_{1\le i\le m}$ none of which are a multiple of the other the following holds for infinitely many $n\in \mathbb{N}$
$$ \varphi (a _1 n+b _1) > \varphi (a _2 n+b _2) > \cdots >\varphi (a _m n+b _m) $$
And besides that there are solutions to $\varphi(2n)=\varphi(2n+1)$ (but it is unknown if there is infinitely many of them. Here is the relevant sequence) so I don't think there is a simple lower bound to $$\left|\frac{1}{\varphi(2n)}-\frac{1}{\varphi(2n+1)}\right|$$
If you want an estimate or an average then it is a different matter.
A: Gjergji is right about the reference to my paper - but I wanted to point out that the specific fact that is needed, namely that inequalities of the form $\phi(an+b) > \phi(cn+d)$ hold infinitely often (provided $ad\ne bc$), was proved earlier:
D. J. Newman, Euler’s $\phi$ function on arithmetic progressions, Amer. Math. Monthly 104 (1997), no. 3, 256–257.
The proof essentially follows Peter and Bjorn's outline, if I remember correctly.
A: See here
A: As a matter of fact your sum diverges, a little manipulation shows that
$$\sum_{n \leq X} \frac{(-1)^n}{\phi(n)} = \sum_{n \leq X, 2|n} \frac{1}{\phi(n)} -
\sum_{n \leq X, (n,2)=1} \frac{1}{\phi(n)} = \sum_{n \leq X/2} \frac{1}{\phi(2n)}
-\sum_{n \leq X,(n,2)=1}\frac{1}{\phi(n)}$$
The above equala to
$$\sum_{n \leq X/2, (n,2)=1} \frac{1}{\phi(2n)} + \sum_{n \leq X/2, 2|n} \frac{1}{\phi(2n)}
 - \sum_{n \leq X, (n,2)=1} \frac{1}{\phi(n)}$$ By multiplicativity of $\phi(n)$ we have $\phi(2n) = \phi(n)$ when $(n,2)=1$. Thus the first sum above is a sum over $1/\phi(n)$ and the above equation simplifies to
$$ - \sum_{X/2 < n \leq X, (n,2)=1} \frac{1}{\phi(n)} + \sum_{n \leq X/4} \frac{1}{\phi(4n)}$$ It follows that
$$\sum_{n \leq X} \frac{(-1)^n}{\phi(n)}= -\sum_{X/2 < n \leq X, (n,2)=1} \frac{1}{\phi(n)} + \sum_{n \leq X/4} \frac{1}{\phi(4n)} \sim c \cdot \log{X}$$
because the first sum on the right converges to a constant, while the second sum on the right is asymptotically $c \cdot \log{X}$. 
EDIT: Put details, erased mention of an earlier confusion about $(-1)^{n+1}$ not being a multiplicative function :P (it is!)
