5. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n}=\frac{1}{4 a_{n-1}}+\frac{1}{n}(n \geqslant 2)$, then the value of $a_{100}$ is $\qquad$ .
$a_{1}=1, a_{2}=\frac{3}{4}, a_{3}=\frac{2}{3}, a_{4}=\frac{5}{8}, a_{2}=\frac{3}{5}, a_{6}=\frac{7}{12}$, conjecture $a_{n}=\frac{n+1}{2 n}$.
If the conclusion holds for $n=k$, then $a_{k+1}=\frac{1}{4 a_{k}}+\frac{1}{k+1}=\frac{k}{2(k+1)}+\frac{1}{k+1}=\frac{k+2}{2(k+1)}$,
which means the conclusion also holds for $n=k+1$. Therefore, $a_{100}=\frac{101}{200}$.