# Determine $\lim_{n \to \infty} a_n$

Given $$a,b\in\mathbb R$$ , defined $$(a_n)_n$$ recursively by setting

$$a_1 = a$$ ,$$a_2 = b$$ ,$$a_{n+1} = \frac{1}{2n}a_{n-1}+\frac{2n-1}{2n}a_n,\;\;n\geq 2$$

Determine $$\displaystyle\lim_{n \to \infty} a_n$$

My attempt : I know that $$a_n = a + (b-a) +.......+ (a_n - a_{n-1})$$ after that I'm not able to procede further.

any hints/solution will be appreciated

Thank you!

• May I ask where you found that task? (: Dec 28 '19 at 17:56

Hint: $$a_{n+1} - a_n = - \frac{1}{2n}(a_n - a_{n-1})$$

Thus $$a_{n+1}-a_n = \frac{(-1)^{n-1}}{(2n)!!}(b-a)$$

$$a_n - a = \sum_{i=1}^{n-1}(a_{i+1}- a_{i}) = (a - b) \sum_{i=1}^{n-1} \frac{(-1)^i}{(2i)!!}$$

The summation certainly converges, but I don't know how to get the converged value

According to wolframalpha, (and thanks to @Minus One Twelfth hint) $$\sum_{i=1}^{\infty} \frac{(-1)^i}{(2i)!!} = \sum_{i=0}^{\infty} \frac{1}{i!}\left(-\frac{1}{2}\right)^i - 1 = e^{-1/2} - 1$$

In summary, $$\lim_{n \rightarrow \infty} a_n = \left(1-\frac{1}{\sqrt{e}}\right) b + \frac{1}{\sqrt{e}}a$$

• Why from this we'll get a convergence? Mar 1 '19 at 7:37
• By the way, for the infinite sum, we can use the fact that $\color{blue}{(2i)!! = 2^{i}i!}$ and then use the Taylor series for $e^{x}$. Mar 1 '19 at 8:10
• Thanks @MinusOne-Twelfth, that solves the summation analytically Mar 1 '19 at 8:21
• @MoonKnight Can we prove $a_{n+1}-a_n = \frac{(-1)^{n-1}}{(2n)!!}(b-a)$ by descending induction? That way I got $\frac{1}{2n\cdot2(n-1)\cdot2(n-2)\cdot\ldots}$ when adding all the terms which is the pattern for the Taylor series. Dec 28 '19 at 18:56