We know the sum of first $n$ natural numbers, their squares and cubes. sum of higher powers can be worked out using the differences: $k^m-(k-1)^{m}$. However, these formulas are not remembered well. Recently, Dr. Mythili Subramanian and I have started wondering if one can write $$S_{n,m}=\sum_{k=1}^{n} k^m =\sum_{j=0}^{m-1} A_{n,j}(m)~ S_{n,j} $$ then what are the expression/name for the coefficients: $A_{n,j}(m)?$
Interestingly, we know the asymptotic result that $$\sum_{k=1}^{n} k^m \sim \frac{n^{m+1}}{m+1}, ~\text{when $n$ is large}.$$ Any suggestion, information or help is welcome here. We are also trying to get it.
If the $A_{n,j}$ are allowed to depend on $m$ as well, one can proceed as follows. We start with $$ k^m = \sum\limits_{j = 0}^{m - 1} {( - 1)^{m + j - 1} \binom{m}{j}S_{k,j} } . $$ Summing over $k$ gives \begin{align*} S_{n,m} & = \sum\limits_{j = 0}^{m - 1} {( - 1)^{m + j - 1} \binom{m}{j}\sum\limits_{k = 1}^n {S_{k,j} } } \\ & = \sum\limits_{j = 0}^{m - 1} {( - 1)^{m + j - 1} \binom{m}{j}(n + 1)S_{n,j} } - \sum\limits_{j = 0}^{m - 1} {( - 1)^{m + j - 1} \binom{m}{j} S_{n,j + 1} } \\ & = - mS_{n,m} + ( - 1)^{m - 1} n(n + 1) + \sum\limits_{j = 1}^{m - 1} {( - 1)^{m + j - 1} \left[ \binom{m}{j}(n + 1) + \binom{m}{j-1} \right]S_{n,j} } . \end{align*} Thus, $$ S_{n,m} = \sum\limits_{j = 0}^{m - 1} {( - 1)^{m + j - 1} \frac{1}{{m + 1}}\left[ \binom{m}{j}(n + 1) + \binom{m}{j-1} \right]S_{n,j} } . $$