I have no idea about how to deal with point B. Can anyone help me? Also, an elegant way to solve point A would be great but it's not that important. Thanks in advance for the help!
A) Suppose $a\in\mathbb R$ with $0 \lt a \lt 1$. Determine: $$\lim_{n\to \infty}\frac{a^n+{a^-}^n}{1+a^n}$$
B) Suppose that: $$a_{n+1}=\frac{-5a_n}{2n+1}$$ for $n=1,2,...$ and $a_1=1.$ Find an expression for $a_{n+1}$ in terms of $n$ and show that $a_{n+1}\rightarrow 0$ as $n \rightarrow \infty$
One may write, for $k=1,2,\cdots,$ $$ \frac{a_{k+1}}{a_k}=-5 \cdot \frac{1}{2k+1} \tag1 $$ giving, for $n=1,2,\cdots,$ $$ \frac{a_{n+1}}{a_n}\times \frac{a_{n}}{a_{n-1}}\times\cdots \times\frac{a_{2}}{a_1}=(-5)^{n}\frac{1}{2n+1} \times \frac{1}{2n-1}\times \cdots \times\frac{1}{2\cdot 1+1} $$ or, by telescoping, $$ \frac{a_{n+1}}{a_1}=(-5)^n\frac{1}{(2n+1)!!} $$ that is
Using Stirling's formula as $n \to \infty$, we get that $$ \frac{(2n)!}{2^{2n}n!^2}\sim\frac1{\sqrt{\pi n}}\tag3 $$ yielding $$ \left|a_n\right|=5^{n-1}\frac{2^{n}n!}{(2n)!}\sim \frac{5^{n-1}}{2^n\cdot n!}\cdot\sqrt{\pi n} \sim \frac1{5\sqrt{2}}\cdot \left(\frac{5e}{2n} \right)^n \to 0 $$ as announced.