I need some help with the harmonic series

[SouthernPotato]

My gratitude to anyone who can tell me or show me where I can find the 32 billionth term of the harmonic series. It should work out to about 24.8, but I need precision.

Alternatively, if you can find the term at which the harmonic series adds to 24.8, that would work as well.

For those unaware, the harmonic series is (insert sigma here) 1/n where n goes from 1 to infinity. I need to know THE VALUE OF N AND THE TOTAL AT THAT TIME.

Thank you.

[Benjio]

Would you not just take the integral from .1 to 32 billion of 1/n? The value of n being 32 billion and the total at that time being 26.4916.

[SouthernPotato]

I need more precision...Also, it shouldn't exceed 24.8 (before I said 28.4, but it was wrong)

[Benjio]

I need more precision...Also, it shouldn't exceed 24.8 (before I said 28.4, but it was wrong)

If you run an integral from 1 (I used .1 earlier) to 32 billion with 1/n you get an answer of 24.189 which doesn't exceed 24.8.

[SouthernPotato]

I could be wrong, but the point of it being a sigma is that it uses integers because it adds one each time (I think). So it starts 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 and so on to infinity. If 32 billion doesn't get to 24.8, what number does?

[Benjio]

I could be wrong, but the point of it being a sigma is that it uses integers because it adds one each time (I think). So it starts 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 and so on to infinity. If 32 billion doesn't get to 24.8, what number does?

Just so you know you're going to have to quote people's posts if you want them to get notifications And to answer your question you will need to set up an integral from 1 to n of 1/x and set the integral equal to 24.8. Once you solve for n you get an answer of 58952625459.8, which also happens to be e^24.8. I'm not sure if you've learnt integrals or not yet, they are the easiest way to solve this problem though.

[SouthernPotato]

Just so you know you're going to have to quote people's posts if you want them to get notifications And to answer your question you will need to set up an integral from 1 to n of 1/x and set the integral equal to 24.8. Once you solve for n you get an answer of 58952625459.8, which also happens to be e^24.8. I'm not sure if you've learnt integrals or not yet, they are the easiest way to solve this problem though.

I understand integrals more in theory than in practice... But doesn't sigma notation mean you're modulating n by 1 each time? Isn't that different than an integral?

Thanks for the tip. I normally quote, really I do, but I was on my phone so it didn't occur to me...

[Benjio]

I understand integrals more in theory than in practice... But doesn't sigma notation mean you're modulating n by 1 each time? Isn't that different than an integral?

Thanks for the tip. I normally quote, really I do, but I was on my phone so it didn't occur to me...

Oh yeah I think I must've been doing it wrong then, sorry  :unsure:  

[TideRower]

What do you know about the harmonic series?

The integration method mentioned above (integrating 1/x from 1 to n+1) gives you an approximation of the value.

If n goes towards infinity the difference converges towards 0.57... (Euler-Mascheroni constant).

Since the result of the integration is ln (n+1)=24.1890 this is the fastes way to "calculate" nth harmonic number.

There doesn't exist a formula that gives you the exact value. You'd have to calculate the whole sum.

 

Here is the result from the last method in double precision : 24.766217404182605.

Edited March 25, 2014 by TideRower

[SouthernPotato]

What do you know about the harmonic series?

The integration method mentioned above (integrating 1/x from 1 to n+1) gives you an approximation of the value.

If n goes towards infinity the difference converges towards 0.57... (Euler-Mascheroni constant).

Since the result of the integration is ln (n+1)=24.1890 this is the fastes way to "calculate" nth harmonic number.

There doesn't exist a formula that gives you the exact value. You'd have to calculate the whole sum.

 

Here is the result from the last method in double precision : 24.766217404182605.

What you're saying is that this is calculated not by using ln, but actually by lengthy summations 32 billion times? If so, that's precisely what I'm after, I believe.

[TideRower]

What you're saying is that this is calculated not by using ln, but actually by lengthy summations 32 billion times? If so, that's precisely what I'm after, I believe.

Yes the second number is the result of the summation (which can be easily done with a pc). I did it in double precision (around 16 precise decimal digits), so it should be the exact number.