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<<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
[tex]S_n = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...[/tex]
to show that the harmonic series diverges. They specifically want this done using the test where
if |an| ≥ dn of a known divergent series, then the series for an also diverges.
In trying to do this, I rearranged the terms in Sn 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:
[tex]\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}} + ...[/tex]
Is that true? I know the limits as n approaches infinity all go to zero.
Here are the details of my attempt at this:
[tex]S_n = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...[/tex]
[tex]= 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}) +...[/tex]
Then I rearranged them like this (since addition is commutative):
[tex]= 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}+ ... +[/tex]
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:
[tex]= \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}} + ...[/tex]
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:
[tex]\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}[/tex]
and since k=n is fine since it's a dummy variable, it just equals:
[tex]\sum_{n=1}^\infty \frac{1}{2^n}[/tex].
Since 1/2n< |1/n|, it then follows that 1/n diverges.
I appreciate you reading this. Please rip it apart. :)
I found a problem in Boas 3rd ed that asks the reader to use
[tex]S_n = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...[/tex]
to show that the harmonic series diverges. They specifically want this done using the test where
if |an| ≥ dn of a known divergent series, then the series for an also diverges.
In trying to do this, I rearranged the terms in Sn 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:
[tex]\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}} + ...[/tex]
Is that true? I know the limits as n approaches infinity all go to zero.
Here are the details of my attempt at this:
[tex]S_n = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...[/tex]
[tex]= 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}) +...[/tex]
Then I rearranged them like this (since addition is commutative):
[tex]= 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}+ ... +[/tex]
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:
[tex]= \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}} + ...[/tex]
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:
[tex]\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}[/tex]
and since k=n is fine since it's a dummy variable, it just equals:
[tex]\sum_{n=1}^\infty \frac{1}{2^n}[/tex].
Since 1/2n< |1/n|, it then follows that 1/n diverges.
I appreciate you reading this. Please rip it apart. :)
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