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evene

Member

Registered: 2015-10-18

Posts: 272

Summation Theorems

Given the general form of a series in Geometric Progression

With a=first term, r=common ratio, s=sum of n terms, then

If r be less than 1 and n be infinite.

a,b,c,d, &c are in Harmonic Progression when the reciprocals 1/a,1/b,1/c,1/d,&c are in Arithematic Progression or when a:b::a-b:b-c between any three consecutive terms.

The nth term of the serie is

How would you prove the formulas?

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Summation Theorems

The nth term of the serie should be

I think the proof is too elementary and no help should be sought from the forum.

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

zetafunc

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Registered: 2014-05-21

Posts: 2,432

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Re: Summation Theorems

Given the general form of a series in Geometric Progression

With a=first term, r=common ratio, s=sum of n terms, then

If r be less than 1 and n be infinite.

More precisely, |r| < 1.

Try calling the sum of the first series S and then multiplying it by r. Then re-arrange to get a formula for S. What happens as n gets large, and why is the assumption |r| < 1 necessary?

Last edited by zetafunc (2016-11-18 22:02:08)

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Summation Theorems

No thickhead, the book wrote it as

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Summation Theorems

O.K. I may be wrong.

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Summation Theorems

hi evene,
Can you check the nth term by substituting n=1,2,3 etc.?

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}