- Thread starter
- #1

- Thread starter bincybn
- Start date

- Thread starter
- #1

- Moderator
- #2

- Feb 7, 2012

- 2,786

Yes, that is true. You can find a proof here.Hii All,

$ \sum_{i=1}^{x}i^{N}:N>2 $. Is there any approximated lower bound for the above summation? Is it > $ \frac{1}{N+1}x^{(N+1)}$ ? If yes, how to prove that?

- Feb 13, 2012

- 1,704

Don't know how to proof it

Setting $\displaystyle S_{n,k}= \sum_{j = 1}^{n} j^{k}$ a fundamental identity is...

$\displaystyle \binom{k+1}{1}\ S_{n,1} + \binom{k+1}{2}\ S_{n,2} + ... + \binom{k+1}{k}\ S_{n,k} = (n+1)\ \{(n+1)^{k}-1\}$ (1)

... and (1) can be written as...

$\displaystyle S_{n,k} = \frac{(n+1)\ \{(n+1)^{k}-1\} - \binom{k+1}{1}\ S_{n,1} - \binom{k+1}{2}\ S_{n,2} - ... - \binom{k+1}{k-1}\ S_{n,k-1}}{\binom{k+1}{k} }$ (2)

... i.e. $S_{n,k}$ is the sum of the term $\displaystyle \frac{(n+1)\ \{(n+1)^{k}-1\}} {\binom{k+1}{k}}$ and a linear combination of the $S_{n,1}$,$S_{n,2}$, ..., $S_{n,k-1}$. That means that $\displaystyle S_{n,1}= \frac{n\ (n+1)}{2}$, which divides the term $\displaystyle \frac{(n+1)\ \{(n+1)^{k}-1\}} {\binom{k+1}{k}}$, divides also all the $S_{n,k}$ for k>1...

Kind regards

$\chi$ $\sigma$

- Thread starter
- #4