Easiest way to calculate the Sum of Harmonic serie?

[Horse]

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?

 

[Triangulum]

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

 

[Horse]

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?

 

[FedEx]

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

 

[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...

 

[Count Iblis]

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"

 

[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

 

[FedEx]

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

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

 

[Horse]

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)}

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?

 

[FedEx]

<br /> 30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{r}{r+1}

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

<br /> 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

 

[Horse]

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.

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

<br /> 30 + 30/31 = \sum_{1}^{30}\frac{2}{r} + \frac{r - 1}{r} - \frac{1}{r + 1}

<br /> 30 + 30/31 = \sum_{1}^{30}\frac{r + 1}{r} - \frac{1}{r + 1}

Now try doing the sum

<br /> 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.

 

[FedEx]

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

 

[FedEx]

http://ecademy.agnesscott.edu/~lriddle/series/rear.htm

Well i think i have bitten of more than what i can chew... I have tried every way which is taught to a 12th grader... but in vain... Just have a look at the link

Count Iblis has rather given a very ingenious solution