series proof q

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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Reply 1

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.

Reply 2

pleasedtobeatyou

1/n^2 converges to 0?

If you're talking about summation, then the value is (pi^2)/6

Reply 3

Felix Felicis

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.

Posted from TSR Mobile

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.

(edited 10 years ago)

Reply 4

ztibor

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

(edited 10 years ago)

Reply 5

cooldudeman

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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Reply 6

ztibor

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