# Easiest way to calculate the Sum of Harmonic serie?

• Horse
In summary, the conversation is about calculating the sum of a series, specifically \sum^{30}_{i=1} 1/i. The initial approach involved breaking the problem into parts and calculating the sum of geometric series. However, this method was deemed to be error-prone and the question was raised if there was an easier way to calculate the sum of harmonic series. One suggestion was to use integration, but there were concerns about its applicability to series. Another approach was suggested using the Euler-Maclaurin summation formula, which involves manipulating the series and taking a limit to get a simpler formula. Ultimately, Count Iblis provided a more efficient solution using a technique involving partial fraction decomposition.

## Homework Statement

Calculate the sum of the serie.

## Homework Equations

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

## 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?

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

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

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?

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

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

Cause i think that its wrong

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...

Horse said:
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?

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

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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:

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

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

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

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FedEx said:
well this is quite a good sum.

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

Now covert the sigma term on the left from r=1 to r=30( right now its 1 to 29)

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

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

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

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

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

FedEx said:
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...

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

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

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

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

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

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

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.

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

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

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

FedEx said:
Now try doing the sum

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

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

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