What do people mean when they say that Euler treated infinity differently? I read in various books that, today, mathematicians would not approve of Euler's methods and his proofs lacked rigor. Can anyone elaborate?
Edit: If I remember correctly Euler's original solution to the Basel problem is as follows.
Using Taylor series for $\sin (s)/s$ we write $\sin (s)/s = 1 - {s^2}/3! + {s^4}/5! - \cdots $ but $\sin (s)/s$ vanishes at $\pm \pi$, $\pm 2\pi$, etc. hence $\frac{{\sin s}}{s} = {\left( {1 - \frac{s}{\pi }} \right)}{\left( {1 + \frac{s}{\pi }} \right)}{\left( {1 - \frac{s}{{2\pi }}} \right)}{\left( {1 + \frac{s}{{2\pi }}} \right)}{\left( {1 - \frac{s}{{3\pi }}} \right)}{\left( {1 + \frac{s}{{3\pi }}} \right)} \cdots$ or $\frac{{\sin s}}{s} = {\left( {1 - \frac{{{s^2}}}{{{1^2}\pi^2}}} \right)}{\left( {1 - \frac{{{s^2}}}{{{2^2}{\pi ^2}}}} \right)}{\left( {1 - \frac{{{s^2}}}{{{3^2}{\pi ^2}}}} \right)} \cdots$ which is $\frac{{\sin s}}{s} = 1 - \frac{{{s^2}}}{{{\pi ^2}}}{\left( {\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \cdots } \right)} + \cdots.$ Equating coefficients yields $\zeta (2) = \frac{{{\pi ^2}}}{6}.$
But $\pm \pi$, $\pm 2\pi$, etc. are also roots of ${e^s}\sin (s)/s$, correct? So equating coefficients does not give ${\pi ^2}/6$.