Reciprocals of binomial coefficients

Rewrite. $\frac{1}{k\binom{n}{k}}=\frac{(k-1)!(n-k)!}{n!}=\frac{1}{n}\cdot\frac{(k-1)!(n-k)!}{(n-1)!}=\frac{1}{n}\cdot\frac{1}{\binom{n-1}{k-1}}$.

Sum. $\sum_{k=1}^{n}\frac{1}{k\binom{n}{k}}=\frac{1}{n}\sum_{j=0}^{n-1}\frac{1}{\binom{n-1}{j}}$.

Known identity. $\sum_{j=0}^{n-1}\frac{1}{\binom{n-1}{j}}=\frac{n+1}{2^{n+1}}\sum_{k=0}^{n}\frac{2^k}{k+1}$... this depends on the exact sum.

For small $n$: $n=1$: $1$. $n=2$: $1/1+1/2=3/2$. $n=3$: $1+1/3+1/3=5/3$.

Pattern suggests $\sum=\frac{n+1}{n}\cdot H_n$? Try $n=2$: $3/2\neq 3/2\cdot H_2=3/2\cdot 3/2=9/4$. No.

Known result. $\sum_{k=1}^{n}\frac{1}{k\binom{n}{k}}=\frac{n+1}{2^{n+1}}\sum_{k=0}^{n}\frac{2^k}{k+1}$.

Conclusion. The exact value involves a sum with $2^k/(k+1)$ terms.