The Recursion Is Trapped In The Exponent!

We have $\left(\frac{2n+1}{2n-1}\right)^{a_n} = e^2 \\ a_n = \frac{2}{\ln\left(\frac{2n+1}{2n-1}\right)}.$ Substituting in the limit, we obtain $\lim_{n \to \infty} \frac{2a_n}{n+1} = \lim_{n \to \infty} \frac{\frac{4}{n+1}}{\ln\left(\frac{2n+1}{2n-1}\right)} \\ =\lim_{n \to \infty} \frac{\frac{-4}{(n+1)^2}}{\frac{-4}{(2n+1)(2n-1)}} \\ =\lim_{n \to \infty} \frac{(2n+1)(2n-1)}{(n+1)^2} \\ 4.$

This problem can be solved in a much easier way. We know that $\underset { y\rightarrow \infty }{ lim } { \left( 1+\frac { 1 }{ y } \right) }^{ 2y }={ e }^{ 2 }$ . Also we observe that $\left( \frac { 2x+1 }{ 2x-1 } \right) =\left( 1+\frac { 2 }{ 2x-1 } \right)$ . Thus we can say $\\ \underset { \frac { 2x+1 }{ 2 } \rightarrow \infty }{ lim } { \left( 1+\frac { 2 }{ 2x+1 } \right) }^{ 2\times \left( \frac { 2x+1 }{ 2 } \right) }\\ =\underset { \frac { 2x+1 }{ 2 } \rightarrow \infty }{ lim } { \left( 1+\frac { 2 }{ 2x+1 } \right) }^{ 2x+1 }\\ ={ e }^{ 2 }$

Now let us look at the given equation. Comparing the above equation with the given one, the following form automatically comes to our mind: ${ a }_{ n }=2n+1,\forall n\in N$ . Then obviously we have: $\underset { n\rightarrow \infty }{ lim } \left( \frac { { 2a }_{ n } }{ n+1 } \right) \\ =\underset { n\rightarrow \infty }{ lim } \left( \frac { 4n+2 }{ n+1 } \right) \\ =4$ The idea behind this proof was that since the question was demanding for a unique answer, so if we can find any sequence that satisfies the given criterion, that sequence will do. This is obviously not a proper way to solve the given problem since it assumes that a unique answer to the given question exists, but still, this is just a trick that can be used in solving multiple-choice type questions.