# Easiest way to calculate the Sum of Harmonic serie?

1. May 9, 2009

### Horse

1. The problem statement, all variables and given/known data

Calculate the sum of the serie.

2. Relevant equations

$$\sum^{30}_{i=1} 1/i$$

3. The attempt at a solution

I broke the problem into parts. I calculated the sum of the geometric series such as:

$$\sum^{4}_{i=1} 2^{-i}$$
$$\sum^{3}_{i=1} 3^{-i}$$
$$\sum^{2}_{i=1} 5^{-i}$$
$$\sum^{1}_{i=1} 7^{-i}$$
and so on for each base with a prime number. Then, I need to sum them to other terms. However, it can be very error-some.

The reason to the question is that I sense that there may be an easier way to calculate the sum. What are the ways to calculate harmonic series? Is my way wrong?

2. May 9, 2009

### Triangulum

You can use integration by integrating 1/x and making the upper limit 30 and the lower limit 1.

3. May 10, 2009

### Horse

It is a serie, not a continuous function. I have never used integration for series. How does its integration differ from the intregation over a continuous function? Is it an approximated value?

4. May 10, 2009

### FedEx

Jesus Christ!!! Where did you learn this? Are you sure about your method..

Cause i think that its wrong

5. May 10, 2009

### FedEx

well this is quite a good sum.

Express 1/2 as 1-1/2, 1/3 as 1-2/3 and so on...

So you will get like

$$30 -\sum_{1}^{29}\frac{r}{r+1} = \sum_{1}^{30}\frac{1}{r}$$

Now covert the sigma term on the left from r=1 to r=30( right now its 1 to 29) Once done with that. Take the sigma term on RHS so you will get rsquare upon r+1.... and some other term...
Now convert rsqare to rsquare-1+1 and simplify... So on rhs you will r and 1 upon r+1 under sigma and some other term...

Now take all the terms from rhs to lhs except the 1 upon r+1... Thats it...

6. May 10, 2009

### Count Iblis

http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula" [Broken]

Last edited by a moderator: May 4, 2017
7. May 10, 2009

### Count Iblis

You can apply the Euler-Maclaurin summation formula by using that:

$$\lim_{L\rightarrow \infty}\sum_{n=1}^{L}\frac{1}{n} - \log(L)=\gamma$$

where $\gamma=0.577215664902\cdots$ is Euler's constant.

What we need to do is avoid using the lower limit of n = 1 in the Euler-Maclaurin summation formula as the series will not converge well. We can write:

$$\sum_{n=1}^{R}\frac{1}{n} = \sum_{n=1}^{L}\frac{1}{n}-\sum_{n=R+1}^{L}\frac{1}{n}$$

Where L is larger than R+1 and is arbitrary. Applying the Euler-Maclaurin summation formula to the summation from R+1 to L gives:

$$\sum_{n=1}^{R}\frac{1}{n}=\sum_{n=1}^{L}\frac{1}{n}-\log(L) +\mathcal{O}\left(L^{{-1}}\right) + \log(R+1) - \frac{1}{2(R+1)}-\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{1}{(R+1)^{2k}}$$

Talking the limit for L to infinity gives:

$$\sum_{n=1}^{R}\frac{1}{n}=\log(R+1) + \gamma - \frac{1}{2(R+1)}-\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{1}{(R+1)^{2k}}$$

If we write:

$$\sum_{n=1}^{R}\frac{1}{n}= \frac{1}{R} + \sum_{n=1}^{R-1}\frac{1}{n}$$

and apply the summation formula to the summation to R-1, we get a nicer looking formula:

$$\sum_{n=1}^{R}\frac{1}{n}=\log(R) + \gamma + \frac{1}{2R}-\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{1}{R^{2k}}$$

For R = 30, we only need the first 3 terms of the summation over the Bernoulli numbers to get a result accurate to within 10^(-14). Writing out these first few terms explicitly gives the formula:

$$\sum_{n=1}^{R}\frac{1}{n}=\log(R) + \gamma + \frac{1}{2R}-\frac{1}{12 R^2}+\frac{1}{120 R^4}-\frac{1}{252 R^6}+\cdots$$

Last edited: May 10, 2009
8. May 10, 2009

### FedEx

well there is no need to have such complicated solution... Ofcourse its damn right but the method which i have shown is equally correct(thats what i feel).... And we may be able to do it with more ease

Last edited by a moderator: May 4, 2017
9. May 10, 2009

### Horse

$$30 + 30/31 -\sum_{1}^{30}\frac{r}{r+1} = \sum_{1}^{30}\frac{1}{r}$$

$$30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{r}{r+1}$$

$$30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{r+1-1}{r+1}$$

$$30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + 1 - \frac{1}{r+1}$$

$$30/31 = \sum_{1}^{30}\frac{1}{r(r+1)}$$

It is hard to follow. Am I on the right track?

10. May 10, 2009

### FedEx

$$30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{r}{r+1}$$

$$30 + 30/31 = \sum_{1}^{30}\frac{r+1+r^2}{(r)(r+1)}$$

$$30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{1}{(r)(r+1)} + \frac{r^2 - 1}{(r)(r+1)}$$

Now convert the middle term to 1 upon r+1 minus 1 upon r. And the last term would get converted to r-1 upon r.

Now try doing the sum

11. May 11, 2009

### Horse

$$30 + 30/31 = \sum_{1}^{30}\frac{1}{r} - (\frac{1}{r+1} - \frac{1}{r}) + \frac{r - 1}{r}$$

$$30 + 30/31 = \sum_{1}^{30}\frac{2}{r} + \frac{r - 1}{r} - \frac{1}{r + 1}$$

$$30 + 30/31 = \sum_{1}^{30}\frac{r + 1}{r} - \frac{1}{r + 1}$$

$$30 + 30/31 = \sum_{1}^{30}\frac{(r + 1)^2 - r}{r(r+1)}$$

It is hard to see where the thing is aiming at.

12. May 11, 2009

### FedEx

seems that i have made some mistake... I am working on it though

13. May 11, 2009