series proof q Watch

cooldudeman
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#1
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how do I do this? would I start with stating that 1/n2 series converges to 3 (according to my notes) then start another proof.

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IceKidd
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it might help if you show us the proof given in your lecture notes, as you are clearly meant to generalise that for cases which work which should be obvious to you due to certain steps in the proof.
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pleasedtobeatyou
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1/n^2 converges to 0?

If you're talking about summation, then the value is (pi^2)/6
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Felix Felicis
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(Original post by cooldudeman)
how do I do this? would I start with stating that 1/n2 series converges to 3 (according to my notes) then start another proof.

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What was the proof of the convergence of \sum_{n\geq 1} 1/n^2 in your lecture notes? I presume it was the integral comparison test. Think about how you can generalise the values of \alpha such that \sum_{n\geq 1} 1/n^{\alpha} converges.
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ztibor
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(Original post by cooldudeman)
how do I do this? would I start with stating that 1/n2 series converges to 3 (according to my notes) then start another proof.

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use the theorem that the series of
\displaystyle \sum_{k=1}^{\infty} a_k is convergent iif
\displaystyle \sum_{l=1}^{\infty} a_{2^l} \cdot 2^l is convergent where terms of (a_k) are positive and (a_k) is monotone decreasing

So
\displaystyle \sum_{k=1}^{\infty} \frac{1}{n^{\alpha}} and
\displaystyle \sum_{l=1}^{\infty} \frac{1}{(2^{\alpha})^l}\cdot 2^l
are convergent and divergent simultaneously.

\displaystyle \sum_{l=1}^{\infty} \frac{1}{(2^{\alpha})^l}\cdot 2^l =\sum_{l=1}^{\infty} \frac{1}{2^{\alpha \cdot l}}\cdot \frac{1}{ 2^{-l}}=
\displaystyle =\sum_{l=1}^{\infty} \left (\frac{1}{2}\right )^{\alpha \cdot l - l}=\sum_{l=1}^{\infty} \left (\frac{1}{2}\right )^{(\alpha -1) \cdot l}=\sum_{l=1}^{\infty} q^l

which is a geometric series and convergent if
|q]=\left (\frac{1}{2}\right )^{\alpha -1} <1 \rightarrow (\alpha -1)>0 \rightarrow \alpha>1
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cooldudeman
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(Original post by IceKidd)
it might help if you show us the proof given in your lecture notes, as you are clearly meant to generalise that for cases which work which should be obvious to you due to certain steps in the proof.
(Original post by Felix Felicis)
What was the proof of the convergence of \sum_{n\geq 1} 1/n^2 in your lecture notes? I presume it was the integral comparison test. Think about how you can generalise the values of \alpha such that \sum_{n\geq 1} 1/n^{\alpha} converges.
yeah sorry I thought it was a general result.
(Original post by ztibor)
use the theorem that the series of
\displaystyle \sum_{k=1}^{\infty} a_k is convergent iif
\displaystyle \sum_{l=1}^{\infty} a_{2^l} \cdot 2^l is convergent where terms of (a_k) are positive and (a_k) is monotone decreasing

So
\displaystyle \sum_{k=1}^{\infty} \frac{1}{n^{\alpha}} and
\displaystyle \sum_{l=1}^{\infty} \frac{1}{(2^{\alpha})^l}\cdot 2^l
are convergent and divergent simultaneously.

\displaystyle \sum_{l=1}^{\infty} \frac{1}{(2^{\alpha})^l}\cdot 2^l =\sum_{l=1}^{\infty} \frac{1}{2^{\alpha \cdot l}}\cdot \frac{1}{ 2^{-l}}=
\displaystyle =\sum_{l=1}^{\infty} \left (\frac{1}{2}\right )^{\alpha \cdot l - l}=\sum_{l=1}^{\infty} \left (\frac{1}{2}\right )^{(\alpha -1) \cdot l}=\sum_{l=1}^{\infty} q^l

which is a geometric series and convergent if
|q]=\left (\frac{1}{2}\right )^{\alpha -1} <1 \rightarrow (\alpha -1)>0 \rightarrow \alpha>1
wow I never knew about that theorem. is there any other way to do it because I haven't even seen that theorem in my notes.

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ztibor
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(Original post by cooldudeman)

wow I never knew about that theorem. is there any other way to do it because I haven't even seen that theorem in my notes.

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If you haven't seen that theorem in your notes, this does not mean that
the theorem does not exist.
This is a proved convergence test named Cauchy condensation test for infinite
series, which you can use for your proof, or
you can begin to prove that.
- For this work: consider that the part sum of a_k will be lower or upper
bound for the part sum of a_(2^l) *2^l depending on choosing value of l
Some help here: http://en.wikipedia.org/wiki/Cauchy_condensation_test
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