Limits

e ??

{}\lim_{n\rightarrow \infty} \left(\frac{a_{1}+1}{a_{1}} \right)\left(\frac{a_{2}+1}{a_{2}} \right)...................\left(\frac{a_{n}+1}{a_{n} }\right)

a_{n-1} +1=\frac{a_{n}}{n}..........................i

\lim_{n\rightarrow \infty } \left(\frac{a_{2}}{2} \right)\left(\frac{a_{3}}{3} \right)\left(\frac{a_{4}}{4} \right)..........\left(\frac{a_{n+1}}{n+1} \right)\frac{1}{a_{1}a_{2}a_{3}.......a_{n}}

\lim_{n\rightarrow \infty } \frac{a_{n+1}}{a_{1}(n+1)!} =\frac{1}{a_{1}}\lim_{n\rightarrow \infty }\frac{1+a_{n}}{n!}.........from i

\frac{1}{a_{1}}\lim_{n\rightarrow \infty }\left(\frac{1}{n!} +\frac{a_{n}}{n!}\right) .................put. n+1.in. place. of. n.in...... a_{n-1} +1=\frac{a_{n}}{n} ...we.get... a_{n} +1 =\frac{a_{n+1}}{n+1}

\frac{1}{a_{1}}\lim_{n\rightarrow \infty }\left(\frac{1}{n!} + \frac{1}{(n-1)!} + \frac{a_{n-1}}{(n-1)!}\right)..........from i

similarly doing it we get

\frac{1}{2} \lim_{n\infty }\left(\frac{1}{n!} + \frac{1}{(n-1)!} + \frac{1}{(n-2)!} ..............\frac{1}{2!} +\frac{1}{1!} +\frac{a_{1}}{1!}\right)

\frac{1}{2} \lim_{n\infty }\left(\frac{1}{n!} + \frac{1}{(n-1)!} + \frac{1}{(n-2)!} ..............\frac{1}{2!} +\frac{1}{1!} +\frac{2}{1!}\right)

\frac{1}{2} \lim_{n\infty }\left(\frac{1}{n!} + \frac{1}{(n-1)!} + \frac{1}{(n-2)!} ..............\frac{1}{2!} +\frac{1}{1!} +1 + 1\right)

since e=\left(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} .......\infty \right)

hence this limit reduces to
=\frac{e+1}{2}

But this answer is not in the options...
But it seems correct.

@tusshar thats wat i hav done naa......
@ injenjoe jus check wat is the value of a1 given ..........bec i hav done an exactly similar q....but in that only diff was that a1 was 1 so ans was coming out to be to be e