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thedarktiger

Member

Registered: 2014-01-10

Posts: 91

Infintite sieries

1) Let S be the sum of a finite geometric series with negative common ratio whose first and last terms are 1 and 4, respectively. (For example, one such series is 1-2+4, whose sum is 3.)

There is a real number L such that S must be greater than L, but we can make S as close as we wish to L by choosing the number of terms in the series appropriately. Determine L.

2)If  a/b rounded to the nearest trillionth is 0.008012018027, where a and b are positive integers, what is the smallest possible value of a+b?/

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Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Infintite sieries

hi

Q1.  The nth term is given by

and the sum by

So  if the nth is also the last,  use these two equations to write S in terms of r

So now consider the graph of r against S.

If you are unsure what this looks like, here it is:

http://www.mathsisfun.com/data/function … x-1)/(x-1)

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Infintite sieries

Q2.

If a number a/b has the value k plus or minus delta then

So figure out what K delta and b are and you have the limits for a + b. 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Infintite sieries

1) Let S be the sum of a finite geometric series with negative common ratio whose first and last terms are 1 and 4, respectively. (For example, one such series is 1-2+4, whose sum is 3.)

There is a real number L such that S must be greater than L, but we can make S as close as we wish to L by choosing the number of terms in the series appropriately. Determine L.

The series clearly has an odd number of terms. Hence

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And since n is odd,

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Thus

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Here I cheated and used WolframAlpha. It turns out that S_n is a decreasing sequence converging to 5/2. Hence

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[*]

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Bassaricyon neblina

thedarktiger

Member

Registered: 2014-01-10

Posts: 91

Re: Infintite sieries

Thanks for the help guys!

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Good. You can read.