Question Number 154275 by mathdanisur last updated on 16/Sep/21
Commented by benhamimed last updated on 16/Sep/21
$$=\Pi\frac{\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({n}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{2}{n}\right)^{\mathrm{2}} }=\Pi\frac{\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{2}\right)\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{2}\right)} \\ $$$$=\Pi\left(\frac{{n}^{\mathrm{2}} +\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\right)\left(\frac{{n}^{\mathrm{2}} +\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$={lim}\left(\frac{\mathrm{2}×\mathrm{5}×\mathrm{10}×..\left({n}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{1}×\mathrm{2}×\mathrm{5}×..\left(\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right)}\right)\left(\frac{\mathrm{2}×\mathrm{5}×\mathrm{10}×…\left({n}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{5}×\mathrm{10}×…×\left(\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right)}\right) \\ $$$$={lim}\left({n}^{\mathrm{2}} +\mathrm{1}\right).\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}=\mathrm{2} \\ $$