Sum of series: using 1 + 1/2 + 1/2 +.... to show 1/n diverges

[Battlemage!]

<<Moderator's note: moved from a technical forum, so homework template missing.>>

I found a problem in Boas 3rd ed that asks the reader to use
S_n = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...

to show that the harmonic series diverges. They specifically want this done using the test where
if |a_n| ≥ d_n of a known divergent series, then the series for a_n also diverges.

In trying to do this, I rearranged the terms in S_n so that they made smaller series, all added together, with each one smaller than 1/n. In order for this to work in the way I'm trying to do it, the following has to be true:

\sum_{n=1}^\infty \frac{1}{2^n} = \sum_{n=1}^\infty \frac{1}{2^{n+1}} + \sum_{n=1}^\infty \frac{1}{2^{n+2}} + \sum_{n=1}^\infty \frac{1}{2^{n+3}} + ...

Is that true? I know the limits as n approaches infinity all go to zero.

Here are the details of my attempt at this:

S_n = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...
= 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}) +...

Then I rearranged them like this (since addition is commutative):

= 1 + \frac{1}{2} + \frac{1}{4}+\frac{1}{8} +...+ \frac{1}{4} + \frac{1}{8}+\frac{1}{16} +...+ \frac{1}{8} + \frac{1}{16}+\frac{1}{32}+ ... +

and so on (the first 1 was giving me fits until I realized 1 = 1/2 + 1/2 anyway, so from this point on I ignored it).This all should then equal:

= \sum_{n=1}^\infty \frac{1}{2^n} + \sum_{n=1}^\infty \frac{1}{2^{n+1}} + \sum_{n=1}^\infty \frac{1}{2^{n+2}} + ...

Where essentially it becomes a double sum with n going to infinity and then m going to infinity (i.e. the exponent is n, n+1, n+2, ... n+m). It's here that it occurred to me that n+m can just be k, and it's all a dummy variable anyway, which would mean that:

\sum_{n=1}^\infty \frac{1}{2^n} + \sum_{n=1}^\infty \frac{1}{2^{n+1}} + \sum_{n=1}^\infty \frac{1}{2^{n+2}} + ... = \sum_{n=1}^\infty \frac{1}{2^k}

and since k=n is fine since it's a dummy variable, it just equals:

\sum_{n=1}^\infty \frac{1}{2^n}.

Since 1/2ⁿ< |1/n|, it then follows that 1/n diverges.
I appreciate you reading this. Please rip it apart. :)

 

[mfb]

You did a lot of unnecessary work, and as result of that unnecessary work the proof does not work any more.

\sum_{n=1}^\infty \frac{1}{2^n} = \sum_{n=1}^\infty \frac{1}{2^{n+1}} + \sum_{n=1}^\infty \frac{1}{2^{n+2}} + \sum_{n=1}^\infty \frac{1}{2^{n+3}} + ...

This equation is true, but the left sum has a finite value. You want to compare the harmonic series to a divergent series, not a convergent one.

Then I rearranged them like this (since addition is commutative):

A finite number of exchanges is okay, but rearrangements like you did do not work - they can change the limit of the series. In particular, with this step you made a convergent series out of your divergent one. Where is the point in everything that follows?

S_n = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...
= 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}) +...

This is the key point, and all you need in terms of series manipulation. The series is clearly divergent, and in the second line every term is smaller than the corresponding term of the harmonic series.

 

[andrewkirk]

As mfb points out, the manipulations you've done break the rules of what an infinite sum means, so they are not valid deductions in classical mathematics. If you're interested in playing around with series though, and what happens if we relax some of the rules, you may be interested in Ramanujan summation, which is pretty topical given a movie just came out recently about Ramanujan.

The concept of a Ramanujan summation can be made rigorous, but it describes an operation that is different from what is accepted in mathematics as calculating the sum of a divergent infinite series. Not only can Ramanujan derive a finite 'sum' of a divergent series but it also can derive different 'sums' by different rearrangements of the order.

 

[Erland]

Since this a series with positive terms, its terms may be rearranged in an arbitrary manner, without changing the sum (which is considered to be $+\infty$ if the sum diverges.

In this case however, you begin with a divergent series and end up with a convergent one:

S_n = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...
= 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}) +...

Then I rearranged them like this (since addition is commutative):

= 1 + \frac{1}{2} + \frac{1}{4}+\frac{1}{8} +...+ \frac{1}{4} + \frac{1}{8}+\frac{1}{16} +...+ \frac{1}{8} + \frac{1}{16}+\frac{1}{32}+ ... +

and so on

and the latter series in convergent, you find correctly its sum as equal to $1+\sum_{n=1}^\infty\frac1{2^n}$ which is a convergent geometric series, not a divergent series.

Your main mistake is that you did not make a true rearrangement, but left out many terms: in your first sum, you have one $\frac12$-term, two $\frac 14$-terms, four $\frac18$-terms, eight $\frac1{16}$-terms etc. while in the second (nested) series you have one $\frac12$-tern, two $\frac14$-terms, three $\frac18$-terms, four $\frac1{16}$-terms, etc.
So, you left out lots and lots of terms. Therefore, it is not sursprising that the first series diverges and the second converges.

 

[SammyS]

I don't see anywhere in this thread where the harmonic series is given. Look it up on Wikipedia or whatever.

Here it is: $\ \displaystyle \sum _{n=1}^{\infty }\frac {1}{n}=1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+{\frac {1}{5}}+\cdots$

 

[Battlemage!]

Since this a series with positive terms, its terms may be rearranged in an arbitrary manner, without changing the sum (which is considered to be $+\infty$ if the sum diverges.

In this case however, you begin with a divergent series and end up with a convergent one:and the latter series in convergent, you find correctly its sum as equal to $1+\sum_{n=1}^\infty\frac1{2^n}$ which is a convergent geometric series, not a divergent series.

Your main mistake is that you did not make a true rearrangement, but left out many terms: in your first sum, you have one $\frac12$-term, two $\frac 14$-terms, four $\frac18$-terms, eight $\frac1{16}$-terms etc. while in the second (nested) series you have one $\frac12$-tern, two $\frac14$-terms, three $\frac18$-terms, four $\frac1{16}$-terms, etc.
So, you left out lots and lots of terms. Therefore, it is not sursprising that the first series diverges and the second converges.

Thanks! That definitely explains it in a very understandable way. My train of thought was that at some point along the road to infinity, each of those groupings should eventually add up to the preceding fraction. Of course that seems to be mistaken. But I can definitely see where I clearly missed terms as the series progressed.

I don't see anywhere in this thread where the harmonic series is given. Look it up on Wikipedia or whatever.

Here it is: $\ \displaystyle \sum _{n=1}^{\infty }\frac {1}{n}=1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+{\frac {1}{5}}+\cdots$

It was given (sloppily) in the last line of the OP:

Since 1/2ⁿ< |1/n|, it then follows that 1/n diverges.