How was the harmonic number's relation to the hurwitz zeta function derived?

Question

How was the generalized harmonics number's relation to the hurwitz zeta function derived?

$H_{n,\ m} =\zeta ( m,\ 1) -\zeta ( m,\ n+1),\ \Re(m)>1$

I tried looking at the series representations for both functions, but I couldn't see how they could relate.

$\sum ^{n}_{k=1}\frac{1}{k^{m}} =\sum ^{\infty }_{k=1}\left[\frac{1}{k^{m}} -\frac{1}{( k+n)^{m}}\right]$

Somehow I just can't find a way to show these two series are equal.

I thought to loook at the integral representations, but I couldn't find the representation for the generalized harmonic numbers.

Any help in proving that both series are equal or in providing a derivation for the relation would be greatly appreciated!

Answer 1

Start with Gamma function

$$\Gamma(m)=\int_0^\infty t^{m-1}e^{-t}dt,\quad \Re(m)>1$$

and by subbing $t=-k\ln(x)$ we have

$$\frac1{k^m}=\frac{(-1)^{m-1}}{\Gamma(m)}\int_0^1 x^{k-1}\ln^{m-1}(x)dx,\quad \Re(m)>1$$

sum up both sides over $n$ we have

$$\sum_{k=1}^n\frac1{k^m}=H_n^{(m)}=\frac{(-1)^{m-1}}{\Gamma(m)}\int_0^1\ln^{m-1}(x)\left(\sum_{k=1}^n x^{k-1}\right)dx.$$

Using the geometric series for the inner sum we obtain

$$H_n^{(m)}=\frac{(-1)^{m-1}}{\Gamma(m)}\int_0^1 \frac{\ln^{m-1}(x)(1-x^n)}{1-x}dx$$

By writing $\frac1{1-x}=\sum_{k=1}^\infty x^{k-1}$ we have

$$H_n^{(m)}=\frac{(-1)^{m-1}}{\Gamma(m)}\sum_{k=1}^\infty\int_0^1\ln^{m-1}(x)[x^{k-1}-k^{n+k-1}]dx$$

$$=\frac{(-1)^{m-1}}{\Gamma(m)}\sum_{k=1}^\infty\left(\frac{(-1)^{m-1}\Gamma(m)}{k^m}-\frac{(-1)^{m-1}\Gamma(m)}{(n+k)^m}\right)$$

$$=\sum_{k=1}^\infty\frac{1}{k^m}-\sum_{k=1}^\infty\frac{1}{(n+k)^m}=\zeta(m)-\sum_{k=1}^\infty\frac{1}{(n+k)^m}$$

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Different approach:

$$H_n^{(m)}=\sum_{k=1}^n\frac{1}{k^m}=\sum_{k=1}^\infty\frac{1}{k^m}-\sum_{k=n+1}^\infty\frac{1}{k^m}$$

reindex the second sum we have

$$H_n^{(m)}=\sum_{k=1}^\infty\frac{1}{k^m}-\sum_{k=1}^\infty\frac{1}{(n+k)^m}$$

But the problem with this proof is that its valid for only integer $n$ as you can see that from manipulating the sum limits.

Answer 2

As @mertunsal commented, for large values of $n$ $$H_n^{(m)}=n^{-m} \left(-\frac{n}{m-1}+\frac{1}{2}-\frac{m}{12 n}+\frac{m^3+3 m^2+2 m}{720 n^3}+O\left(\frac{1}{n^4}\right)\right)+\zeta (m)$$ while $$\zeta (m,1)-\zeta (m,n+1)=\zeta (m)-\zeta (m,n+1)=\text{ the same}$$

Answer 3

This is just a telescoping.

E.g.

$$\frac1{1^3}+\frac1{2^3}+\frac1{3^3}+\frac1{4^3}\\=\\\left(\frac1{1^3}-\frac1{5^3}\right)+\left(\frac1{2^3}-\frac1{6^3}\right)+\left(\frac1{3^3}-\frac1{7^3}\right)+\left(\frac1{4^3}-\frac1{8^3}\right)+\left(\frac1{5^3}-\frac1{9^3}\right)+\cdots$$