# Harmonic series partial sum

According to Beyer (1987) the following progression is called harmonic series: $$\frac {1} {a_{1}},\frac {1} {a_{1}+d}, \frac {1} {a_{1}+2d},...$$

How it can be calculated the partial sum of the above mentioned sequence?

• I don´t see that there exists a closed form for the partial sum. – callculus Nov 17 '18 at 16:03
• @callculus what about approximation? – David Nov 17 '18 at 16:03
• Search for "Harmonic numbers". Note however, that your sequence is more general, related to "Digamma function" – Yuriy S Nov 17 '18 at 16:06
• @YuriyS Thank you for your comment. Can you please provide some material reagrdinf this problem? – David Nov 17 '18 at 16:10

The sum of terms for $$n=0,1,...,m$$ is $$\sum_{n=0}^m \frac{1}{d(a_1/d+n)} =\frac{1}{d} \left(\psi \left(m+1+\frac{a_1}{d}\right) -\psi \left(\frac{a_1}{d}\right) \right)$$

Where $$\psi$$ is the Digamma function.

Use the definition:

$$\psi(z)=-\gamma+\sum_{n=0}^\infty \frac{z-1}{(n+1)(n+z)}$$

• Thank you for your answer. Could you please provide some information how it is derived? P.S. sorry, if my comment is weird, i don't have background in this topic. – David Nov 17 '18 at 16:23
• @David, try using the definition I provided, and you will see that the formula is true – Yuriy S Nov 17 '18 at 16:25
• Thank you. My aim is to find a closed-form expression for the finite series. Is it possible? – David Nov 17 '18 at 16:31
• @David, I have written it in my answer above – Yuriy S Nov 17 '18 at 16:38
• The Digamma function is expressed in terms of infinite sum, but my aim is to find closed-form expression. Thank you. – David Nov 17 '18 at 16:52

Yuriy S gave the only possible closed form for the summation.

For large $$n$$, for sure, you could use series expansions and get from $$S_n=\sum_{i=0}^n \frac{1}{a+i\,d}=\frac 1d \left(\psi \left(n+1+\frac{a}{d}\right)-\psi \left(\frac{a}{d}\right) \right)$$ $$S_n=\frac{\log(n)-\psi \left(\frac{a}{d}\right)}{d}+\frac{2 a+d}{2 d^2 n}-\frac{6 a^2+6 a d+d^2}{12 d^3 n^2}+\frac{2 a^3+3 a^2 d+a d^2}{6 d^4 n^3}+O\left(\frac{1}{n^4}\right)$$

Using $$a=\pi$$, $$d=e$$ and $$n=10$$, the exact result would be $$\approx 1.03107$$ while the above series expansion would give $$\approx 1.03113$$.

• Please have a look. Thank you for your support. math.stackexchange.com/questions/3003278/… – David Nov 18 '18 at 8:56
• @David. No idea about this other one. I am quite skeptical about even an approximation. Even the case $a=\alpha=0$ makes the problem more than difficult (at least to me). – Claude Leibovici Nov 18 '18 at 9:11
• Do you mean both of them? – David Nov 18 '18 at 9:12

The Harmonic Series

The partial sum of the standard Harmonic Series is given by $$H_n=\sum_{k=1}^n\frac1k\tag1$$ This can be extended to a function that is analytic except at the negative integers $$H(z)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z}\right)\tag2$$ The Euler-Maclaurin Sum Formula gives the asymptotic expansion $$H_n=\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}+\frac1{240n^8}-\frac1{132n^{10}}+O\!\left(\frac1{n^{12}}\right)\tag3$$ where $$\gamma=0.57721566490153286060651209$$ is the Euler-Mascheroni Constant.

The Series in the Question \begin{align} \sum_{k=1}^n\frac1{a+(k-1)d} &=\frac1d\sum_{k=1}^n\frac1{\frac ad+k-1}\\ &=\frac1d\sum_{k=1}^\infty\left(\frac1{\frac ad+k-1}-\frac1{\frac{a-d}d+k-1+n}\right)\\[3pt] &=\frac1d\left(H\!\left(n+\frac ad-1\right)-H\!\left(\frac ad-1\right)\right)\tag4 \end{align} using the extension in $$(2)$$.

Formula $$(2)$$ from this answer allows us to compute $$H\!\left(\frac ad-1\right)$$ for rational $$a$$ and $$d$$, while formula $$(3)$$ above allows us to approximate $$H\!\left(n+\frac ad-1\right)$$ for large $$n$$.

• Thank you for your answer. – David Nov 18 '18 at 20:34
• Please have a look. math.stackexchange.com/questions/3003278/… – David Nov 18 '18 at 20:35
• Could you, please, support in order to calculate or approximate the sums represented in the above mentioned link? – David Nov 18 '18 at 20:36