Real series/Inverse of squares/Convergence/Example

We want to determine whether the series

${}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}=1+{\frac {1}{4}}+{\frac {1}{9}}+{\frac {1}{16}}+{\frac {1}{25}}+\ldots \,$

converges. We use the direct comparison test and example, where we have shown the convergence of ${}\sum _{k=1}^{n}{\frac {1}{k(k+1)}}$. For ${}k\geq 2$ we have

${}{\frac {1}{k^{2}}}\leq {\frac {1}{k(k-1)}}\,.$

Hence, ${}\sum _{k=2}^{\infty }{\frac {1}{k^{2}}}$ converges and therefore also ${}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}$. This does not say much about the exact value of the sum. With much more advanced methods, one can show that this sum equals ${}{\frac {\pi ^{2}}{6}}$.