The harmonic function is defined in terms of the Psi and Zeta functions as follows.

FunctionAdvisor(definition, harmonic);

$\left[\mathrm{harmonic}\left(z\right)=\mathrm{\Psi }\left(z+1\right)+\mathrm{\gamma }\,\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\mathrm{harmonic}\left(a\,z\right)=\mathrm{\zeta }\left(z\right)-\mathrm{\zeta }\left(0\,z\,a+1\right)\,\mathrm{with\; no\; restrictions\; on}\left(a\,z\right)\right]$

When the first parameter is a non-negative integer n, the harmonic function admits a Sum representation

FunctionAdvisor(sum_form, harmonic(n));

$\left[\mathrm{harmonic}\left(n\right)=\sum _{\mathrm{\_k1}=1}^n\frac{1}{\mathrm{\_k1}}\&comma;n::\&apos;\mathrm{nonnegint}\&apos;\right],\left[\mathrm{harmonic}\left(n\right)=\sum _{\mathrm{\_k1}=1}^{\mathrm{\infty }}\frac{n}{\mathrm{\_k1}\left(\mathrm{\_k1}+n\right)}\&comma;n::\left(\neg \&apos;\mathrm{negint}\&apos;\right)\right],\left[\mathrm{harmonic}\left(n\right)=\sum _{\mathrm{\_k2}=0}^{\mathrm{\infty }}\sum _{\mathrm{\_k1}=0}^{\mathrm{\infty }}\frac{{\left(\mathrm{-1}\right)}^{\mathrm{\_k1}}{n}^{\mathrm{\_k1}+1}}{{\left(\mathrm{\_k2}+1\right)}^{\mathrm{\_k1}+2}}\&comma;\left|n\right|<1\right]$

FunctionAdvisor(sum_form, harmonic(n,z));

$\left[\mathrm{harmonic}\left(n\&comma;z\right)=\sum _{\mathrm{\_k1}=1}^n\frac{1}{{\mathrm{\_k1}}^z}\&comma;n::\&apos;\mathrm{nonnegint}\&apos;\right],\left[\mathrm{harmonic}\left(n\&comma;z\right)=\left(\sum _{\mathrm{\_k1}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{\_k1}}^z}\right)-\left(\sum _{\mathrm{\_k1}=0}^{\mathrm{\infty }}\frac{1}{{\left(n+1+\mathrm{\_k1}\right)}^z}\right)\&comma;1<\mathrm{\Re }\left(z\right)\right],\left[\mathrm{harmonic}\left(n\&comma;z\right)=\sum _{\mathrm{\_k1}=1}^{\mathrm{\infty }}\left(-\frac{\mathrm{pochhammer}\left(z\&comma;\mathrm{\_k1}\right)\mathrm{\zeta }\left(z+\mathrm{\_k1}\right)n^{\mathrm{\_k1}}{\left(\mathrm{-1}\right)}^{\mathrm{\_k1}}}{\mathrm{\_k1}!}\right)\&comma;\left|n\right|<1\wedge 1<\mathrm{\Re }\left(z\right)\right]$

When the first parameter is a negative integer an exception (error) is raised, signaling the event 'division_by_zero'. This behavior can be controlled using a NumericEventHandler, which will be passed complex infinity as the default value.

When the first parameter is a small non-negative integer and the second parameter, if present, is a non-negative integer, harmonic returns a rational number.

$\mathrm{harmonic}\left(3\right)$

$\frac{11}{6}$

$\mathrm{harmonic}\left(3\&comma;2\right)$

$\frac{49}{36}$

$\mathrm{harmonic}\left(r\&comma;s\right)$

$\mathrm{harmonic}\left(r\&comma;s\right)$

$=\mathrm{convert}\left(\&comma;\mathrm{Sum}\right)\mathbf{assuming}r::\&apos;\mathrm{nonnegint}\&apos;$

$\mathrm{harmonic}\left(r\&comma;s\right)=\sum _{\mathrm{\_k1}=1}^r\frac{1}{{\mathrm{\_k1}}^s}$

$=\mathrm{convert}\left(\&comma;\mathrm{\zeta }\right)$

$\mathrm{harmonic}\left(r\&comma;s\right)=\mathrm{\zeta }\left(s\right)-\mathrm{\zeta }\left(0\&comma;s\&comma;r+1\right)$

$=\mathrm{convert}\left(\&comma;\mathrm{\Psi }\right)\mathbf{assuming}s::\&apos;\mathrm{posint}\&apos;$

$\mathrm{harmonic}\left(r\&comma;s\right)=\frac{{\left(\mathrm{-1}\right)}^s\left(\mathrm{\Psi }\left(-1+s\&comma;1\right)-\mathrm{\Psi }\left(-1+s\&comma;r+1\right)\right)}{\left(-1+s\right)!}$

$\mathrm{diff}\left(\&comma;r\right)$

$s\mathrm{\zeta }\left(0\&comma;s+1\&comma;r+1\right)=-\frac{{\left(\mathrm{-1}\right)}^s\mathrm{\Psi }\left(s\&comma;r+1\right)}{\left(-1+s\right)!}$

$\mathrm{evalf}\left(\mathrm{eval}\left(\&comma;\left[r=\frac{10}{43}+\frac{\mathrm{I}}{2}\&comma;s=4\right]\right)\right)$

$\mathrm{-0.2942981267}-0.9671639794\mathrm{I}=\mathrm{-0.2942981267}-0.9671639794\mathrm{I}$

$\mathrm{FunctionAdvisor}\left(\mathrm{special\_values}\&comma;\mathrm{harmonic}\right)$

$\left[\mathrm{harmonic}\left(0\right)=0\&comma;\mathrm{harmonic}\left(1\right)=1\&comma;\mathrm{harmonic}\left(\mathrm{-1}\right)=\mathrm{\infty }+\mathrm{\infty }\mathrm{I}\&comma;\mathrm{harmonic}\left(\mathrm{\infty }\right)=\mathrm{\infty }\&comma;\mathrm{harmonic}\left(-\mathrm{\infty }\right)=\mathrm{\infty }\&comma;\mathrm{harmonic}\left(0\&comma;z\right)=0\&comma;\mathrm{harmonic}\left(1\&comma;z\right)=1\&comma;\mathrm{harmonic}\left(a\&comma;0\right)=a\&comma;\mathrm{harmonic}\left(a\&comma;1\right)=\mathrm{harmonic}\left(a\right)\&comma;\mathrm{harmonic}\left(\mathrm{-1}\&comma;z\right)=\mathrm{\infty }+\mathrm{\infty }\mathrm{I}\right]$