I am trying to obtain an explicit solution for $p$ from the following equation:
$p = 1 - \left[ \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \frac{1}{k+1} \right]^{\frac{1}{{n}}}$.
Obviously, I know that summing a binomial pdf we have
$\sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} = 1$
Perhaps there is a neat way to rewrite
$\sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \frac{1}{k+1}$
so I get an explicit solution for $p$.
$$ \eqalign{ & \sum\limits_{\left( {0\, \le \,} \right)\,k\,\left( { \le \,n} \right)} {{1 \over {k + 1}}\left( \matrix{ n \cr k \cr} \right)p^{\,k} q^{\,n - k} } = \sum\limits_{0\, \le \,k\,\left( { \le \,n} \right)} {{1 \over {n + 1}}\left( \matrix{ n + 1 \cr k + 1 \cr} \right)p^{\,k} q^{\,n - k} } = \cr & = {1 \over {\left( {n + 1} \right)p}}\sum\limits_{0\, \le \,k\,\left( { \le \,n} \right)} {\left( \matrix{ n + 1 \cr k + 1 \cr} \right)p^{\,k + 1} q^{\,\left( {n + 1} \right) - \left( {k + 1} \right)} } = \cr & = {1 \over {\left( {n + 1} \right)p}}\sum\limits_{1\, \le \,j\,\left( { \le \,n + 1} \right)} {\left( \matrix{ n + 1 \cr j \cr} \right)p^{\,j} q^{\,\left( {n + 1} \right) - j} } = \cr & = {1 \over {\left( {n + 1} \right)p}}\left( {\sum\limits_{0\, \le \,j\,\left( { \le \,n + 1} \right)} {\left( \matrix{ n + 1 \cr j \cr} \right)p^{\,j} q^{\,\left( {n + 1} \right) - j} } - q^{\,\left( {n + 1} \right)} } \right) = \cr & = {{1 - q^{\,n + 1} } \over {\left( {n + 1} \right)p}} = {1 \over {\left( {n + 1} \right)}}{{1 - q^{\,n + 1} } \over {1 - q}} \cr} $$