# Rookie question - Series 1/n

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• March 22nd 2011, 07:15 AM
iva
Rookie question - Series 1/n
Hi there,

I've read through why the sum of the series 1/n ( n=1 to infinity) diverges. Ok but still it is mindboggling because the term of the series will get smaller and smaller surely never being able to reach infinity, yet it is clear that it TENDs to infinity.. just very wierd because I would have thought there could be a point where it could get so tiny that it would maybe converge to some value?

Just interested in hearing a different take on this series compared to my text book and if there is some example in the real world where this series is applied

Thank you
• March 22nd 2011, 08:47 AM
Sambit
As you proceed, each term becomes less than its previous one, but still, all of them are positive. So in spite of the fact that each term decreases, the series effectively gives you a sum of infinite number of positive terms--which can not converge. The individual terms getting smaller and smaller does not imply at all that their sum will converge.
• March 22nd 2011, 08:51 AM
Defunkt
Sambit, I would be careful with that explanation. You could also say that the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ gives you a sum of an infinite number of positive terms - but it does converge.
The point is that for the series to converge, the general term must get close to 0 very fast, otherwise - such as in this case - the series will diverge to infinity.
• March 22nd 2011, 09:10 AM
iva
It makes sense that it diverges, sort of. But it just goes against intuition because the numbers get smaller and smaller so they could never surely add up to infinity . I thought that eventually it would not be able to add up to more than a certain number, but OK :)

Thanks guys
• March 22nd 2011, 09:23 AM
Opalg
There is a good discussion in the Wikipedia article on the harmonic series, including the two standard proofs that the series diverges.
• March 22nd 2011, 09:54 AM
chisigma
Quote:

Originally Posted by iva
... just interested in hearing a different take on this series compared to my text book and if there is some example in the real world where this series is applied...

There is a 'beatiful' problem using the divergent series $1+ \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{2n+1}$ that is known as 'the desert trip'...

... a truck can travel one mile with one fill up of gas. Demonstrate that if at the start of a desert an unlimited quantity of gas is allowable, then the truck can travel along k miles of desert no matter 'how big' is k...

... do You know the solution?...

Kind regards

$\chi$ $\sigma$
• March 22nd 2011, 10:22 AM
iva
I looked that problem up wow that is such a nice example! I'ts set to rest this anti intuitive concept!

Thank you :)

If anyone have more real life examples please post would love to read up on more.. doesn't have to be the same series, any interesting one applied to real situations, biology, anything
• March 22nd 2011, 05:14 PM
Prove It
You can prove the divergence of the harmonic series using the comparison test.

Clearly $\displaystyle \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16} + \dots$

$\displaystyle > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \dots$

$\displaystyle = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots$

$\displaystyle \to \infty$.

Since the harmonic series is greater than this series that goes to $\displaystyle \infty$, then the harmonic series must also go to $\displaystyle \infty$, and get there quicker...
• March 22nd 2011, 09:14 PM
iva
I can see how the 3rd row 1 + 1/2 + 1/2 .. can get there quicker but not the row before it surely? When you say quicker in series don't you have to take term for term ie if you add up the first n variables of either then only you can see which one is quicker surely?
• March 22nd 2011, 09:37 PM
Prove It
The first row is greater than the second row.

The second and third rows are equal, since $\displaystyle \frac{1}{4} + \frac{1}{4} = \frac{1}{2},\,\, \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}$, etc.

So the first row is greater than the third row.
• March 22nd 2011, 10:16 PM
Sambit
Quote:

Originally Posted by Defunkt
Sambit, I would be careful with that explanation. You could also say that the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ gives you a sum of an infinite number of positive terms - but it does converge.
The point is that for the series to converge, the general term must get close to 0 very fast, otherwise - such as in this case - the series will diverge to infinity.

Thanks a lot. This was a very much valid point that I ignored.
• March 22nd 2011, 10:17 PM
iva
it doesn't make sense, in the same essense couldn't i take the first row 1+ 1/2 + 1/3 and slow it down, ie 1 + (1/8 + 1/8 +1/8 +1/8) + (1/12 + 1/12 + 1/12 + 1/12 ).... and then say it grows the slowest?
• March 25th 2011, 09:48 AM
iva
Using harmonic series in a comparison test for convergence:

If i now have a series n^(-1/root n), this series will be less than 1/n as it grows but much slower than the harmonic series right? So can i use the harmonic series to say since 0 < an < bn where an is my series and bn is the harmonic series can I conclude then that my series diverges?
• March 25th 2011, 02:14 PM
awkward
Another pretty application of the harmonic series is the book stacking problem:

Book Stacking Problem -- from Wolfram MathWorld
• March 25th 2011, 06:18 PM
Prove It
Quote:

Originally Posted by iva
Using harmonic series in a comparison test for convergence:

If i now have a series n^(-1/root n), this series will be less than 1/n as it grows but much slower than the harmonic series right? So can i use the harmonic series to say since 0 < an < bn where an is my series and bn is the harmonic series can I conclude then that my series diverges?

No.

The way the comparison test works is:

If you can show that a series is less than a different convergent series, then the original series converges (because anything less than a number is still a number).

OR

If you can show that a series is greater than a different series that diverges to $\displaystyle \infty$, then the original series diverges to $\displaystyle \infty$ (because something greater than $\displaystyle \infty$ is still $\displaystyle \infty$, it just gets there quicker).
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last