Series 1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2 + 1/7... This is the exercise 2.7.2 e) of the book "Understanding Analysis 2nd edition" from Stephen Abbott, and asks to decide wether this series converges or diverges:
$$ 1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots $$
I have noticed that
$$\dfrac{1}{3}< 1-\dfrac{1}{2^2}+\dfrac{1}{3};\\
\dfrac{1}{3}+\dfrac{1}{5}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5};\\
\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}\\
$$
which is true in general because
$$1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{36}-\dfrac{1}{64}-\dots=1-\sum_{n=1}^\infty \dfrac{1}{(2n)^2}=1-\dfrac{1}{4}\sum_{n=1}^\infty\dfrac{1}{n^2}=1-\dfrac{\pi^2}{24}>0.
$$
Thus
$$
\sum_{n=1}^\infty \dfrac{1}{2n+1}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots
$$
Finally, as the series $\sum_{n=1}^\infty \dfrac{1}{2n+1}$ diverges, so does the series requested.
My two questions are:
I) Is this reasoning correct?
II) Can this exercise be done without using that $\sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$? I would like to find a solution with more elementary tools.
Thank you.

Don't be swayed by the particular case. You have the following 
Lemma Let $\sum a_n$ be a series (resp. $\sum b_n$ be a convergent series), then the intertwining $x_{2n}=a_n$ and $x_{2n+1}=b_n$ is convergent iff $\sum a_n$ is so. 
Proof Let us, for short, suppose the indexing starting from zero. Then, we get
$$
\sum_{n=0}^{2N+1}x_n=\sum_{n=0}^N a_n + \sum_{n=0}^N b_n
$$
this proves the equivalence
$$
\sum x_n\mbox{ converges }\Longleftrightarrow \sum a_n\mbox{ converges }
$$
end of proof
Here
$$ 
1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\ldots 
$$ 
diverges while
$$\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}\ldots 
$$
converges. Hence your series diverges. 
