Partial sum of the harmonic series

[AwesomeTrains]

Homework Statement

I have to find a natural number N that satisfies this equation:

[itex]\sum^{N}_{i=1} \frac{1}{i} > 100 [/itex]

Homework Equations

I tried finding a close form of the sum but couldn't find anything useful.

The Attempt at a Solution

Well after trying some numbers in maple I found a few very large numbers satisfying the inequality.

Any hints are much appreciated.

 

[Dick]

Homework Statement

I have to find a natural number N that satisfies this equation:

[itex]\sum^{N}_{i=1} \frac{1}{i} > 100 [/itex]

Homework Equations

I tried finding a close form of the sum but couldn't find anything useful.

The Attempt at a Solution

Well after trying some numbers in maple I found a few very large numbers satisfying the inequality.

Any hints are much appreciated.

Think of the sum as representing a Riemann sum approximating the value of an integral.

 

[AwesomeTrains]

Attempt

Thanks for the tip, I have only been doing analysis for around 3 months and we haven't started using integrals yet. Until now we have only looked at sequences and series, are there any other ways of doing it?
Anyways, here is my attempt.
[itex]\sum[/itex][itex]^{N}_{i}\frac{1}{i} > \int ^{N+1}_{1} \frac{1}{i}di = ln(N+1) = 100[/itex]
(From wikipedia)
Then I can easily solve for N and get [itex]N=e^{100}-1\approx 2.6881\cdot10^{43}[/itex]
Thanks for the help :)

 

[Dick]

Thanks for the tip, I have only been doing analysis for around 3 months and we haven't started using integrals yet. Until now we have only looked at sequences and series, are there any other ways of doing it?
Anyways, here is my attempt.
[itex]\sum[/itex][itex]^{N}_{i}\frac{1}{i} > \int ^{N+1}_{1} \frac{1}{i}di = ln(N+1) = 100[/itex]
(From wikipedia)
Then I can easily solve for N and get [itex]N=e^{100}-1\approx 2.6881\cdot10^{43}[/itex]
Thanks for the help :)

Yes, I think that's it. But your N there isn't an integer - you want to pick any integer greater than that number. Just a detail.

 

[AlephZero]

Have you seen the proof that ##\sum_1^\infty \frac 1 i## diverges? You can use that to get a value of N without doing any calculations with logs and exponentials.

The basic idea of the proof is
1/3 + 1/4 > 2(1/4) = 1/2
1/5 + 1/6 + 1/7 + 1/8 > 4(1/8) = 1/2
1/9 ... + 1/16 > 8(1/16) = 1/2
etc

So you can find N = a power of 2, that makes the sum > 200(1/2).