1
$\begingroup$

I'm trying to figure out the closed form formula for the following:

$$ \sum_{i=1}^{N} \frac{1}{i} $$

I think the denominator will have $N!$ in it but otherwise I cannot figure it out. Can someone help?

$\endgroup$
3
$\begingroup$

For a 'closed' form, look to this identity (but, there is no closed, closed form):

$$\text{S}=\sum_{\text{n}=a}^{\text{m}}\frac{1}{n}=\frac{1}{a}+\frac{1}{a+1}+\dots+\frac{1}{\text{m}-1}+\frac{1}{\text{m}}=\psi^{(0)}(\text{m}+1)-\psi^{(0)}(\text{a})$$

Where $\psi^{(n)}(x)$ is the nth derivative of the digamma function.


When $a=1$:

$$\sum_{\text{n}=1}^{\text{m}}\frac{1}{n}=1+\frac{1}{2}+\dots+\frac{1}{\text{m}-1}+\frac{1}{\text{m}}=\psi^{(0)}(\text{m}+1)-\psi^{(0)}(1)=\gamma+\psi^{(0)}(\text{m}+1)=\text{H}_{\text{m}}$$

Where $\text{H}_{\text{x}}$ is the xth harmonic number and $\gamma$ is the Euler-Mascheroni constant.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.