<s>[INST] Limits of the strings 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?
 [/INST] 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 $$\forall~~n~~a_n\le \frac{1}{n}\to 0$$

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