## BdMO National Higher Secondary 2019/6

samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

### BdMO National Higher Secondary 2019/6

When a function $f(x)$ is differentiated $n$ times ,the function we get id denoted $f^n(x)$.If $f(x)=\dfrac {e^x}{x}$.Find the value of
$\lim_{n \to \infty} \dfrac {f^ {2n}(1)}{(2n)!}$

soyeb pervez jim
Posts: 21
Joined: Sat Jan 28, 2017 11:06 pm

### Re: BdMO National Higher Secondary 2019/6

$f(x)=\frac{e^x}{x}=\frac{1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots}{x}=\frac{\sum_{n=0}^{\infty}\frac{x^n}{n!}}{x}=\sum_{n=0}^{\infty}\frac{x^{n-1}}{n!}=\frac{1}{x}+1+\frac{x}{2!}+\frac{x^2}{3!}\dots$
As, $\frac{d^k}{dx^k}(x^q)=\frac{q!}{(q-k)!}x^{q-k}$ (for $q \geq k$) and $\frac{d^k}{dx^k}(x^q)=0$ (for $q<k$)
so, $f^{(2n)}(x)=\frac{2n!}{x^{2n+1}}+0+0+\dots+0+0+\frac{2n!}{(2n+1)!}+\frac{(2n+1)!}{(2n+2)!}x+\frac{(2n+2)!}{2!\times (2n+3)!}x^2+\frac{(2n+3)!}{3!\times (2n+4)!}x^3+\dots$
$=\frac{2n!}{x^{2n+1}}+0+0+\dots+0+0+\sum_{k=0}^{\infty}\frac{x^k}{k!\times(2n+k+1)}$

$\Longrightarrow f^{(2n)}(1)=2n!+0+0+\dots+0+0+\sum_{k=0}^{\infty}\frac{1}{k!\times(2n+k+1)}$

Now,
$\lim_{n\to\infty} \frac{f^{(2n)}(1)}{2n!}=\frac{2n!+0+0+\dots+0+0+\sum_{k=0}^{\infty}\frac{1}{k!\times(2n+k+1)}}{2n!}$
$\Longrightarrow \lim_{n\to\infty} \frac{f^{(2n)}(1)}{2n!}=\frac{2n!}{2n!}+0+0+\dots+0+0+\sum_{k=0}^{\infty}\frac{1}{2n!\times k!\times(2n+k+1)}$
as, $n\to\infty \Longrightarrow \sum_{k=0}^{\infty}\frac{1}{2n!\times k!\times(2n+k+1)}\to 0$
so, $\lim_{n\to\infty} \frac{f^{(2n)}(1)}{2n!}=\frac{2n!}{2n!}+0+0+0+\dots=1$

samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BdMO National Higher Secondary 2019/6

@Jim huge solution.
I also solved it after the olympiad.

I posted the problem @ AoPS.
You can see that solution too.