A partir de la recurrencia
$$a_{n} = 1 + \left(\dfrac{n+1}{2n}\right) a_{n-1} $$
ya tienes
$$a_n=\frac{n+1}{2^n}\sum_{k=1}^{n+1}\frac{2^{k-1}}{k}$$
Esta respuesta muestra una forma de obtener, a partir de aquí,
$$a_n=\sum_{k=0}^{n}\frac{1}{\binom nk}$$
El uso de los siguientes dos lemas (la prueba para el Lema 1 se escribe al final de la respuesta) :
Lema 1 : $\quad\displaystyle\sum_{k=1}^{n+1}\frac{2^{k-1}}{k}=\displaystyle\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{1-(-1)^k}{2k}$
Lema 2 : $\quad (n+1)\displaystyle\int_0^1x^k(1-x)^{n-k}dx=\dfrac{1}{\binom nk}\qquad $ (ver aquí para la prueba)
uno tiene
$$\begin{align}a_n&=\frac{n+1}{2^n}\sum_{k=1}^{n+1}\frac{2^{k-1}}{k}
\\\\&\stackrel{\text{L. %#%#%}}=\frac{n+1}{2^{n}}\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{1-(-1)^k}{2k}
\\\\&=\frac{n+1}{2^{n+2}}\sum_{k=1}^{n+1}\binom{n+1}{k}(1-(-1)^k)\cdot\frac{1-(-1)^k}{k}
\\\\&=\frac{n+1}{2^{n+2}}\sum_{k=1}^{n+1}\binom{n+1}{k}(1-(-1)^k)\int_{-1}^1t^{k-1}dt
\\\\&=\frac{n+1}{2^{n+2}}\int_{-1}^1\sum_{k=1}^{n+1}\binom{n+1}{k}t^{k-1}(1-(-1)^k)dt
\\\\&=\frac{n+1}{2^{n+2}}\int_{-1}^1\frac 1t\sum_{k=0}^{n+1}\binom{n+1}{k}t^k(1-(-1)^k)dt
\\\\&=\frac{n+1}{2^{n+2}}\int_{-1}^1\frac 1t\bigg(\sum_{k=0}^{n+1}\binom{n+1}{k}t^k-\sum_{k=0}^{n+1}\binom{n+1}{k}(-t)^k\bigg)dt
\\\\&=\frac{n+1}{2^{n+2}}\int_{-1}^{1}\frac{(1+t)^{n+1}-(1-t)^{n+1}}{t} dt
\\\\&=(n+1)\int_{1}^{-1}\frac{(1-\frac{1-t}{2})^{n+1}-(\frac{1-t}{2})^{n+1}}{t}\cdot\frac{dt}{-2}
\\\\&=(n+1)\int_0^1\frac{(1-x)^{n+1}-x^{n+1}}{1-2x}dx\qquad\quad \left(\text{set %#%#%}\right)
\\\\&=(n+1)\int_0^1(1-x)^n\cdot\frac{1-\left(\frac{x}{1-x}\right)^{n+1}}{1-\frac{x}{1-x}}dx
\\\\&=(n+1)\int_0^1\sum_{k=0}^{n}x^k(1-x)^{n-k}dx
\\\\&=\sum_{k=0}^{n}(n+1)\int_0^1x^k(1-x)^{n-k}dx
\\\\&\stackrel{\text{L. %#%#%}}=\sum_{k=0}^{n}\frac{1}{\binom nk}\end{align}$$
Por último, vamos a demostrar Lema 1.
Lema 1 : $1$
La prueba de Lema 1 :
Esta es una prueba por inducción.
Que tiene de $\frac{1-t}{2}=x$ desde $2$.
Suponiendo que se tiene para $\quad\displaystyle\sum_{k=1}^{n+1}\frac{2^{k-1}}{k}=\displaystyle\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{1-(-1)^k}{2k}$ y el uso de
$n=0$ (ver aquí para la prueba)$\displaystyle\sum_{k=1}^{0+1}\frac{2^{k-1}}{k}=1=\displaystyle\sum_{k=1}^{0+1}\binom{0+1}{k}\frac{1-(-1)^k}{2k}$
$n$ (ver aquí para la prueba)$\qquad \displaystyle\sum_{k=0}^{n}\frac{1}{k+1}\binom nk=\frac{2^{n+1}-1}{n+1}\qquad$
uno tiene
$$\begin{align}&\sum_{k=1}^{n+2}\binom{n+2}{k}\frac{1-(-1)^k}{2k}-\sum_{k=1}^{n+2}\frac{2^{k-1}}{k}
\\\\&=\frac{1-(-1)^{n+2}}{2(n+2)}+\sum_{k=1}^{n+1}\binom{n+2}{k}\frac{1-(-1)^k}{2k}-\sum_{k=1}^{n+2}\frac{2^{k-1}}{k}
\\\\&=\frac{1-(-1)^{n+2}}{2(n+2)}+\sum_{k=1}^{n+1}\bigg(\binom{n+1}{k}+\binom{n+1}{k-1}\bigg)\frac{1-(-1)^k}{2k}
\\&\qquad\qquad -\bigg(\frac{2^{n+1}}{n+2}+\sum_{k=1}^{n+1}\frac{2^{k-1}}{k}\bigg)
\\\\&=\frac{1-(-1)^{n+2}}{2(n+2)}+\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{1-(-1)^k}{2k}+\sum_{k=1}^{n+1}\binom{n+1}{k-1}\frac{1-(-1)^k}{2k}
\\&\qquad\qquad -\bigg(\frac{2^{n+1}}{n+2}+\sum_{k=1}^{n+1}\frac{2^{k-1}}{k}\bigg)
\\\\&\stackrel{I.H.}=\frac{1-(-1)^{n+2}}{2(n+2)}+\sum_{k=1}^{n+1}\binom{n+1}{k-1}\frac{1-(-1)^k}{2k}-\frac{2^{n+1}}{n+2}
\\\\&=\frac{1-(-1)^{n+2}}{2(n+2)}+\frac 12\sum_{k=1}^{n+1}\frac{1}{k}\binom{n+1}{k-1}-\frac 12\sum_{k=1}^{n+1}\frac{(-1)^k}{k}\binom{n+1}{k-1}-\frac{2^{n+1}}{n+2}
\\\\&=\frac{1-(-1)^{n+2}}{2(n+2)}+\frac 12\bigg(\sum_{k=1}^{n+2}\frac{1}{k}\binom{n+1}{k-1}-\frac{1}{n+2}\binom{n+1}{n+1}\bigg)
\\&\qquad\qquad +\frac 12\bigg(\sum_{k=1}^{n+2}\frac{(-1)^{k-1}}{k}\binom{n+1}{k-1}-\frac{(-1)^{n+1}}{n+2}\binom{n+1}{n+1}\bigg)-\frac{2^{n+1}}{n+2}
\\\\&\color{red}{=}\frac{1-(-1)^{n+2}}{2(n+2)}+\frac 12\bigg(\frac{2^{n+2}-1}{n+2}-\frac{1}{n+2}\bigg)
\\&\qquad\qquad+\frac 12\bigg(\frac{1}{n+2}-\frac{(-1)^{n+1}}{n+2}\bigg)-\frac{2^{n+1}}{n+2}
\\\\&=0\end{align}$$
donde $\qquad\qquad (1)$ fueron utilizados en la igualdad en rojo.$\qquad\displaystyle\sum_{k=0}^{n}\frac{(-1)^k}{k+1}\binom nk=\frac{1}{n+1}\qquad$
