Limits of the strings

Question

What is the limit of this: $a_n=\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}$, where n tends to infinity? I would also like an intuitive explanation in addition to the logical one. Thanks in advance! P.s.: Do you know where can I find the math API on this site?

Answer 1

As n tends to infinity,

$1! \le n!$, $2! \le n!\cdots $, $n! \le n!$

So numerator is upper bounded by $n n!$

Now denominator is 2n! which is $2n*(2n-1)*\ldots * (n!)$

So our equation is upper bounded by $(n*(n!)) / (2n*(2n-1)* \ldots * (n!))$

$n!$ and n cancels out

So we are left with

$1/(2*(2n-1)*(2n-2) . . (n+1))$

As n tends to infinity, it tends to 0.

So, since our upper bounds tends to 0 and our lower bound is 0, hence by sandwhich, the limit on an as n tends to infinity should be 0

QED

Answer 2

$a_n\le \frac{n\cdot n!}{2n\cdot (2n-1)!} = \frac12 \frac{n!}{(2n-1)!}\to0$. Intuitively, $(2n)!$ is huge because you're multiplying numbers all the way to $2n$.

Also: the Math API :) (but where is the emoji API?).

Answer 3

By induction prove that,

$$a_n\le \frac{1}{n}$$

• for $n=1$ $$a_1 = \frac{1}{2}\le \frac{1}{1}$$ Assume that

$$a_n=\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}\le \frac{1}{n} $$

$$a_{n+1}=\dfrac{(1!+2!+\dots+n!)+(n+1)!}{2(n+1)(2n+1)(2 \cdot n)!}\\=\frac{1}{2(n+1)(2n+1)}\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}+\frac{1}{2(n+1)}\dfrac{(n+1)n!}{(2n+1)(2 \cdot n)!}$$ But $$n!\le (2n)!~~~and~~~n+1\le 2n+1 \implies\dfrac{(n+1)n!}{(2n+1)(2 \cdot n)!}\le 1 $$ By assumption $$ \dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}\le\frac{1}{n}\le 1$$ we get

$$ a_{n+1}=\dfrac{(1!+2!+\dots+n!)+(n+1)!}{(2 n+2)!}\le \frac{1}{2(n+1)(2n+1)}+\frac{1}{2(n+1)}\le \frac{1}{n+1}$$

since $$\frac{1}{2n+1}\le 1$$

conclusion $$\forall~~n~~a_n\le \frac{1}{n}\to 0$$