Sum of the series $\sum_{k=0}^n \frac{1}{k+1} \binom{n}{k}$ The task is to transform $\sum_{k=0}^n \frac{1}{k+1} \binom{n}{k}$ into a compact and not recursive formula.
I've done a bunch of similar (as I thought at first) series. I'll describe my method. I name $\sum_{k=0}^n f(k)=a_{n}$, and then describe $a_{n}=a_{n-1}+f(n)$. Then I solve recursive formula by, for example, generating functions. It works for a series like $\sum_{k=0}^n k^2(k+2)$, so a series which doesn't contain $n$ inside.
About the main question, I tried to transform it with some formulas I know, like $\frac{n-k}{k+1}\binom{n}{k}=\binom{n}{k+1}$ or $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$, but none of them worked. With the last one I was able to transform it to like $a_{n}=a_{n+1}+\sum_{0}^{n-1}\frac{1}{k+2}\binom{n-1}{k}$, but it doesn't do anything, because there is $\frac{1}{k+2}$ and not $\frac{1}{k+1}$.
I would appreciate any hints!
Also, I thought about using the series $\sum_{k\geq0} \binom{n}{k}x^{k}=(1+x)^{n}$, because it looks like $\sum_{k=0}^n \binom{n}{k} \frac{1}{k+1}=\sum_{k\geq 0}\binom{n}{k} \frac{1}{k+1}$, and it looks pretty familiar.

Hint:
$$\frac1{k+1}\binom{n}{k}=\frac1{n+1}\binom{n+1}{k+1}$$
