Most natural way to prove $\sum_{n=1}^{\infty}\frac{1}{n+2}$ diverges I don't know how my teacher wants me to prove that 
$$\sum_{n=1}^{\infty}\frac{1}{n+2}$$
diverges. All I know is that I have to use the $a_n>b_n$ criteria and prove that $b_n$ diverges. 
I tried this:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} = 
\frac{1}{1+2}+\frac{1}{2+2}+\frac{1}{3+2} + \cdots + \frac{1}{n+2} = \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left(\frac{1}{1}+\frac{1}{2}\right) = \left(\sum_{n=1}^{\infty}\frac{1}{n}\right)-\frac{3}{2}$$
but I can't get a relationship between $a_n$ and $b_n$ of these series.
If I go to the root of the comparsion criterion, I know that:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} = 
\frac{1}{1+2}+\frac{1}{2+2}+\frac{1}{3+2} + \cdots + \frac{1}{n+2} = \left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots+\frac{1}{n}> \frac{2}{4}+\frac{4}{8}+\cdots+g_n$$ for some $g_n$ that I'm lazy to calculate.
Therefore, by the properties of limits, since the righthand sum $p_n$ diverges, the lefthand sum $s_n$ diverges too, because $s_n>p_n$.  But I don't know if my teacher would accept that, I think she would only accept that I use the argument for the $a_n$ of the sum, not for the partial sum. 
I also thought about rewriting:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} = \sum_{n=3}^{\infty} \frac{1}{n}$$
but I don't see how it helps because the indexes are different.
All I need is an argument that will work with the $a_n$ of the $\sum_{n=1}^{\infty}a_n$, in relation with $b_n$ from $\sum_{n=1}^{\infty} b_n$
 A: $$
\sum_{n=1}^N \frac{1}{n+2} = 
\sum_{n=3}^{N+2} \frac{1}{n}
$$
So your series differs just by finite many terms from the harmonic series, which diverges.
A: You can do in this way:
If $\sum_{n=1}^{\infty} \frac{1}{n+2} $ converges, say $\sum_{n=1}^{\infty} \frac{1}{n+2} = S$.
$\sum_{n=1}^{\infty} \frac{1}{n}=1+\frac12+\sum_{n=1}^{\infty} \frac{1}{n+2}=\frac 32+S$,
Which will contradics with $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges.
A: When you're only trying to distinguish between convergence and divergence, you can toss around big numbers pretty freely:
$$
\sum_{n = 1}^\infty \frac{1}{n+2} \geq \sum_{n = 1}^\infty \frac{1}{1000n} = \frac{1}{1000}\sum_{n = 1}^\infty \frac{1}{n}
$$
The righthand sum diverges ("dividing infinity by 1000 leaves you with infinity"), which implies the lefthand sum diverges ("it is even larger than infinity").
Please note that I do not advocate literally treating infinity as a number, but the statements I use in quotes are useful so far as they go, and probably rigorous statements of them appear in your text.
A: For all $n$, $n + 2 \leq n + 2n$, so 
$$
\sum_{n=1}^\infty \frac{1}{3n} \leq \sum_{n=1}^\infty\frac{1}{n + 2},
$$
and $\sum_{n=1}^\infty \frac{1}{3n}$ diverges since it is a constant multiple of a divergent series. 
