I need to prove the following inequality for those two Summations. However, I am unsure how to tackle this problem :(
$$H_N^{(r)}=\sum_{i=1}^N\frac{1}{i^r}<\sum_{k=0}^{m-1}\frac{2^k}{2^{kr}}. $$
We have :
$N = (2^m)-1$
$r > 1$.
I tried developing both summations :
The H(N) function is the Harmonic serie 1/i^r. H(2^m - 1) = 1/(1^r) + 1/(2^r) + 1/(3^r) ... + 1/(2^m - 1)^r Let G = summation(k=0 to m-1) 2^k / (2^kr) = 1/1 + 2/(2^2r) + 2^2/(2^3r)... + 2^(m-1)/2^(m-1)r
I'm not sure how to proceed from there (and if developing both summations is necessary in solving this).
I thought about comparing terms by terms, but H(N)'s index is bigger ((2^m) -1) > G(m-1), so I don't think that's the correct way of approaching this.
Any suggestions or resources?
Thank you!
Expanding my precedent hint $$ \eqalign{ & {1 \over {2^{\,n} }} + {1 \over {2^{\,n} + 1}} + \cdots + {1 \over {2^{\,n} + \left( {2^{\,n} - 1} \right)}} < 2^{\,n} \left( {{1 \over {2^{\,n} }}} \right) = 1 \cr & \quad \quad \Downarrow \cr & \left( {{1 \over {2^{\,n} }}} \right)^{\,r} + \left( {{1 \over {2^{\,n} + 1}}} \right)^{\,r} + \cdots + \left( {{1 \over {2^{\,n} + \left( {2^{\,n} - 1} \right)}}} \right)^{\,r} < 2^{\,n} \left( {{1 \over {2^{\,n} }}} \right)^{\,r} = {{2^{\,n} } \over {2^{\,r\,n} }} \cr & \quad \quad \Downarrow \cr & \sum\limits_{i = 1}^{2^{\,m} - 1} {{1 \over {i^{\,r} }}} = {1 \over {1^{\,r} }} + \left( {\left( {{1 \over 2}} \right)^{\,r} + \left( {{1 \over {2 + 1}}} \right)^{\,r} } \right) + \cdots + \left( {\left( {{1 \over {2^{\,m - 1} }}} \right)^{\,r} + \left( {{1 \over {2^{\,m - 1} + 1}}} \right)^{\,r} + \cdots + \left( {{1 \over {2^{\,m - 1} + \left( {2^{\,m - 1} - 1} \right)}}} \right)} \right) = \cr & = \sum\limits_{n = 0}^{m - 1} {\sum\limits_{k = 0}^{2^{\,n} - 1} {\left( {{1 \over {2^{\,n} + k}}} \right)^{\,r} } } < \sum\limits_{n = 0}^{m - 1} {{{2^{\,n} } \over {2^{\,r\,n} }}} \cr} $$ for $2 \le m$ (for $m=0,1$ it's an equality).