Identify The Series!

Relevant wiki: Taylor Series - Problem Solving

$a_n=\dfrac{n!}{(n-1)!}\times a_{n-1}$

$~~=\dfrac{n!}{(n-1)!}\times\dfrac{(n-2)!}{a_{n-2}}$

$~~=\dfrac{n!}{(n-1)!}\times\dfrac{(n-2)!}{(n-3)!}\times a_{n-3}$

$\cdots\cdots$

$\begin{cases} =\dfrac{n\cancel !}{\cancel{(n-1)!}}\times\dfrac{(n-2)\cancel{!}}{\cancel{(n-3)!}}\cdots\times 2 \ \color{#D61F06}{\small{\text{if n is even}}}\\=\dfrac{n\cancel !}{\cancel{(n-1)!}}\times\dfrac{(n-2)\cancel{!}}{\cancel{(n-3)!}}\cdots\times 3\times 1\ \color{#D61F06}{ \small{\text{if n is odd}}}\end{cases}$

Hence if n is even,

$a_n=n\times (n-2)\times (n-4)\times 2$

$~~=2^{n/2}\times \left(\dfrac n2\right)!$

Required sum is therefore:

$\therefore \sum_{n=0}^{\infty}\dfrac{1}{2^n \times n!}=\sum_{n=0}^{\infty}\dfrac{\left(\frac 12\right)^n}{n!}=e^{1/2}$

Hence, $\Large 1+2=\boxed 3$

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Alternate Solution:- Above was the actual proof but after writing few terms one can resort to induction to prove:

$a_{2n}=2^n n!, n\in \mathbb{N}$ This is obviously true for $n=1$ since $a_2=2$ . Now assuming it to be true for some natural $k$ i.e $a_{2k}=2^{k}k!$ , we have to prove it also holds for $k+1$ .

Using given recurrence relation , $a_{2k+2}=(2k+2)\times\dfrac{(2k)!}{a_{2k}}=2^{k+1}(k+1)!$

Hence whenever statement is true for some natural $k$ it is also valid for $k+1$ . Hence by induction the statement is true for all natural numbers i.e

$a_{2n}=2^n n!~\forall n\in\mathbb{N}$

And the rest process is same as above.

Relevant wiki: Taylor Series - Problem Solving

By the reccurence we have $a_{1}=1$ $a_{2}=2$ $a_{3}=3$ $a_{4}=2\cdot 4$ $a_{5}=1\cdot 3 \cdot 5$ $a_{6}=2\cdot 4 \cdot 6$ so $a_{2n}=2\cdot 4 \cdot 6 \cdots 2n$ $a_{2n}=2^nn!$ hence $\sum_{n=0}^{\infty}{\frac{1}{a_{2n}}=\sum_{n=0}^{\infty}\frac{(\frac{1}{2})^n}{n!}}=e^{1/2}$ $\alpha + \beta=3$

First few $a_n$ indicates that $a_n = n!!$ . Let us prove it by induction that the claim that $a_n = n!!$ applies to all $n \ge 0$ .

1. For $n=0$ , $a_0 = 0!! = 1$ as given, therefore, the claim is true for $n=0$ .

2. Assuming that the claim is true for $n$ , then we have $a_{n+1} = \dfrac {(n+1)!}{a_n}$ $= \dfrac {(n+1)!}{n!!}$ $= (n+1)!!$ . The claim is also true for $n+1$ , therefore, it is true for all $n \ge 0$ .

Now, we have: $\displaystyle \sum_{n=0}^\infty \frac 1{a_{2n}} = \sum_{n=0}^\infty \frac 1{(2n)!!} = \sum_{n=0}^\infty \frac 1{2^n n!} = e^\frac 12$

$\implies \alpha + \beta = 1+2 = \boxed{3}$