Question

Evaluate the following

${lim}_{n\to \infty }\left\{\right(1+\frac{1}{n^2}\left)\right(1+\frac{2^2}{n^2})\dots (1+\frac{n^2}{n^2}){\}}^{1/n}$

• A. $e^{(\pi -4)/2}$
• B. $2e^{(\pi -4)/2}$
• C. $\frac{e^{(\pi -4)/2}}{2}$
• D. $e^{(\pi -4)}$

Answer 1

The correct option is B $2e^{(\pi -4)/2}$

$y=\underset{n⟶\infty }{lim}{\{(1+\frac{1}{n^2})(1+\frac{2^2}{n^2}).........(1+\frac{n^2}{n^2})\}}^{\frac{1}{n}}y=\underset{n⟶\infty }{lim}{\{\left(\frac{n^2+1}{n^2}\right)\left(\frac{n^2+2^2}{n^2}\right).........\left(\frac{n^2+n^2}{n^2}\right)\}}^{\frac{1}{n}}y=\underset{n⟶\infty }{lim}{\sum }_{1=1}^n{\left(\frac{n^2+1}{n^2}\right)}^{\frac{1}{n}}\log y={\int }_0^1\log (1+x^2)dx$

Using integration by parts

$\log y=[x\log (1+x^2){]}_0^1-{\int }_0^{1}x\frac{2x}{1+x^2}dx\log y=\log 2-2{\int }_0^1\frac{x^2+1-1}{1+x^2}dx\log y=\log 2-2{\int }_0^{1}1-\frac{1}{1+x^2}dx\log y=\log 2-2+2\times \frac{\pi }{4}y={2e}^{\frac{\pi }{4}-2}\Rightarrow y=2e^{(\pi -\frac{4}{2})}$