I am trying to interpret an equation, but can't understand how the less-than signs work:
$\sum _{n=1}^{\infty } \frac{1}{n^2}<1+\sum _{n=1}^{\infty } \frac{1}{n (n+1)}=1+\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{n+1)}\right)=1+1=2$
The first part returns False when I put it into Mathematica which I don't think is how it should be interpreted since this value approximates 1.644.
$\sum _{n=1}^{\infty } \frac{1}{n^2}<1$
On
Since $ \left(\forall n\geq 2\right),\ n^{2}\geq n\left(n-1\right) $, we get that $ \left(\forall n\geq 2\right),\ \frac{1}{n^{2}}\leq\frac{1}{n\left(n-1\right)} $, thus : $$ \sum_{n=1}^{+\infty}{\frac{1}{n^{2}}}=1+\sum_{n=2}^{+\infty}{\frac{1}{n^{2}}}<1+\sum_{n=2}^{+\infty}{\frac{1}{n\left(n-1\right)}}=1+\sum_{n=1}^{+\infty}{\frac{1}{n\left(n+1\right)}} $$
I suppose you can now understand why the first inequality is true.
Your statement "unpacks"
First $$\sum _{n=1}^{\infty } \frac{1}{n^2}<1+\sum _{n=1}^{\infty } \frac{1}{n (n+1)}$$ second $$1+\sum _{n=1}^{\infty } \frac{1}{n (n+1)}=1+\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{n+1)}\right)$$ third $$1+\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{n+1)}\right)=1+1$$ fourth $$1+1=2$$
The use of multiple equalities and inequalities like this is a bit abuse of notation. What if I wrote $$1+1+1=1+2=3$$ then you might be tempted to say $1+1+1=1+2$ is true, so the above becomes $$\text{true}=3$$ (and in certain programming languages the statement really would be interpreted like that) but this is almost never what is meant. What is meant is that the series of (in)equalities should be unpacked, here as $1+1+1=1+2$ and $1+2=3$, independently.