$a_{n+1}=\frac{n}{n+1} a_n$

$a_{n+1} = \frac{n}{n+1}a_n$

$(n+1)a_{n+1} = na_n$

Substituting the values $\displaystyle n\in$ { $\displaystyle 134,133,\ldots,1$ },

$\displaystyle 135\cdot a_{135} = 134\cdot a_{134}$
$\displaystyle 134\cdot a_{134} = 133\cdot a_{133}$
$\displaystyle\vdots$
$\displaystyle 2\cdot a_2 = a_1$

Using all these equations,

$135\cdot a_{135} = a_1$ $135\cdot a_{135} = 5$

$\Rightarrow a_{135} = \boxed{\frac{1}{27}}$

When you solve for $a_{2}$ , $a_{3}$ and $a_{4}$ you see pattern. And that comes out to be $\frac{5}{n}$ . So, $a_{135}$ will be $\frac{5}{135}$ . That comes out to be $\frac{1}{27}$

$a_{n+1}=\frac {n}{n+1}a_{n}$

$a_{n}=\frac {n-1}{n}a_{n-1}$

$a_{n-1}=\frac {n-2}{n-1}a_{n-2}$

$.$

$.$

$\Rightarrow a_{n+1}=\frac {n!}{(n+1)!}a_{1}$

$\Rightarrow a_{n+1}=\frac {1}{n+1}a_{1}$

now substituting $n+1=135$

$\boxed{a_{135}=\frac {1}{27}}$