Note, also, the working on the second page can be substantially simplified, in my opinion. Starting from
}}} \qquad \text{. . . (*)})
and then simplifying through the use of log laws and cancelling like terms, rather than extracting the telescoping series. Consider the summation term:
}}&=\sum_{n=2}^k{\ln{\left(\cfrac{1}{1-\frac{1}{n^2}}\times\cfrac{n^2}{n^2}\right)}}=\sum_{n=2}^k{\ln{\left[\frac{n^2}{(n+1)(n-1)}\right]}}\\&=\ln{\left(\frac{2^2}{3\times 1}\right)}+\ln{\left(\frac{3^2}{4\times 2}\right)}+\ln{\left(\frac{4^2}{3\times 5}\right)}+...+\ln{\left[\frac{(k-1)^2}{k(k-1)}\right]}+\ln{\left[\frac{k^2}{(k+1)(k-1)}\right]}\\&=\ln{\left[\frac{2^2\times 3^2\times 4^2\times ...\times (k-1)^2\times k^2}{1\times 2\times3^2\times 4^2\times 5^2\times ...\times (k - 1)^2\times k\times(k+1)}\right]}\quad\text{(log laws)}\\&=\ln{\left[\frac{2^2}{1\times 2}\times\frac{1}{1}\times\frac{k^2}{k(k+1)}\right]}\\&=\ln{\left(\frac{2k}{k+1}\right)}\\ \\ \text{And,}\qquad\lim_{k\to\infty}{\ln{\left(\frac{2k}{k+1}\right)}}&=\lim_{k\to\infty}{\ln{\left(\frac{2k}{k+1}\times\cfrac{\frac{1}{k}}{\frac{1}{k}}\right)}}=\lim_{k\to\infty}{\ln{\left(\cfrac{2}{1+\frac{1}{k}}\right)}}=\ln{2}\quad\text{as $\frac{1}{k}\to0$}\end{align*})
}}} &= \lim_{k\to\infty}{\ln{\left(\frac{2k}{k+1}\right)}} = \ln{2} \quad \text{into (*) yields:} \\ 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... &< 1 + \ln{2} \qquad \text{as required} \end{align*})
and then simplifying through the use of log laws and cancelling like terms, rather than extracting the telescoping series. Consider the summation term: