In the book ‘Riemann’s Zeta Function’ by H.M Edwards, the following is a line in a proof (within section 6.7) that I can’t follow. The variable $T$ is just a positive real number.
$$|\Re [\zeta(2+iT)]|\geq 1 - 2^{-2}-3^{-2}-4^{-2}-… = 1 - (\zeta(2)-1)$$
Where does this result come from?
This comes from the fact that $$\Re\zeta(2+iT)=\Re\sum_{n=1}^\infty \frac{1}{n^{2+iT}}=\sum_{n=1}^\infty \frac{\Re n^{-iT}}{n^2}.$$ Since $n^{-iT}=e^{-iT\log n}=\cos(T\log n)-i\sin(T\log n)$, the numerator of each term is between $-1$ and $1$. The numerator of the $n=1$ term is $1$, so we have $$\Re\zeta(2+iT)\geq 1-\frac1{2^2}-\frac1{3^2}-\frac1{4^2}-\cdots.$$