# sum of harmonic progression?

Someone asked me for a formula for the sum of the harmonic progression. So I did some calculations and gave him an approximate formula:

$$\int_1^n\frac{dx}{x} = \frac{y_1 + y_2}{2} + \frac{y_2 + y_3}{2} + \cdots +\frac{y_{n-1} + y_n}{2}$$ where $y_i$ is $i$th term of the HP $$\ln(n) = \frac{y_1}{2} + y_2 +y_3 + \cdots +\frac{y_n}{2}$$

so

$$\sum_{i=1}^n y_i = \ln(n) + \frac{y_1 + y_n}{2}$$

e.g. $1+1/2+\cdots+1/10 = 2.8525$

actual result $= 2.9289$

My question is, how to correct this formula?

The partial sums of the harmonic series are called "harmonic numbers." The difference between the nth harmonic number and ln(n) tends to a limit as n increases, and that limit is called Euler's constant or gamma.

There's a great book about all this called Gamma: Exploring Euler's Constant.

A good quick approximation: $\log(n + 0.5) + \gamma$.

A yet more accurate approximation: $\log(n)+\gamma+1/2n + 1/12n^2-1/120n^4+1/252n^6-1/240n^8+1/132n^{10}-691/32760n^{12}+\cdots$, taking as many terms as is convenient.

Error analysis:

The first formula above has maximum error of $1/24n^2$. uday's first answer has error of about 0.0024 ln n. Mine is a better approximation for all $n\neq6$.

The first formula is also better than uday's revised formula for all $n\ge46$. The expected error in uday's answer is about 0.000016.

Hey guys.... There is no correct simple general formula for sum to n terms of the series 1+1/2+1/3+1/4+ ............. + 1/n

but the following formula will be a good approximation for sum to n-terms of the above series when n>5

S = log(n+0.5) + 0.5772 + 0.04021/(n*n + 0.8848)

Deviation from the actual value fluctuates but remains relatively low..

so i guess this may be a good approximation

• it would be interesting to see how you found these numbers if they are not just experimentally determined.
– anon
Oct 13, 2010 at 17:55
• @muad: To get close to the best approximation with constants in those places, 1.841 should be about $\exp(\gamma)$, 0.9973 should be about 1, and 0.6184 should be about 1/2. Oct 13, 2010 at 18:19
• @muad: Hi... i have taken values of n from 1 to 1000000 and found the values of S. Using curvefitting tool in matlab i came to that formula...
– uday kiran
Oct 20, 2010 at 11:21
• @charls: Hi... i totally agree with you but for smaller values of n,log(n+0.5)+γ has significant amount of error.. so for about n=30 my formula will give a better approximation...
– uday kiran
Oct 20, 2010 at 11:25
• @uday kiran: Your new formula is worse than my simple formula for all n > 45. Oct 24, 2010 at 23:55

The harmonic progression has a simple and elegant formula, though it's arguably not a closed-form. If $$a$$ and $$b$$ are integers:

$$\sum_{j=1}^{n}\frac{1}{a j+b}=-\frac{1}{2b}+\frac{1}{2(a n+b)}-\pi\int_{0}^{1}(1-u)(\cos{2\pi(a n+b)u}-\cos{2\pi b u})\cot{\pi a u}\,du$$

We can generalize this formula, which in the case of odd powers is given by: $$\begin{multline} \sum_{j=1}^{n}\frac{1}{(a j+b)^{2k+1}}=-\frac{1}{2b^{2k+1}}+\frac{1}{2(a n+b)^{2k+1}}+\\ -\frac{(-1)^{k}(2\pi)^{2k+1}}{2}\int_{0}^{1}\sum_{j=0}^{k}\frac{B_{2j}\left(2-2^{2j}\right)(1-u)^{2k+1-2j}}{(2j)!(2k+1-2j)!}\left(\cos{2\pi(a n+b)u}-\cos{2\pi bu}\right)\cot{\pi au}\,du \end{multline}$$

Since there is more than one way to derive these formulas, another possibility is below ($$i$$ is the imaginary unit): $$\begin{multline} \sum_{j=1}^{n}\frac{1}{(a j+b)^{k}}=-\frac{1}{2b^{k}}+\frac{1}{2(a n+b)^{k}}\\+\frac{i(2\pi\,i)^{k}}{2}\int_{0}^{1}\sum_{j=0}^{k}\frac{B_{j}(1-u)^{k-j}}{j!(k-j)!}\left(e^{2\pi i(a n+b)u}-e^{2\pi i\,bu}\right)\cot{\pi a u}\,du \end{multline}$$

The demonstration on how one goes about deriving these formulas was given in this paper.