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#1 2015-01-19 17:48:40

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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#2 2015-01-19 21:47:56

Bob
Administrator
Registered: 2010-06-20
Posts: 10,010

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


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 smile

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#3 2015-01-19 22:47:32

Bob
Administrator
Registered: 2010-06-20
Posts: 10,010

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.  smile

Bob


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 smile

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#4 2015-01-21 01:25:10

Olinguito
Member
Registered: 2014-08-12
Posts: 649

Re: Infintite sieries

thedarktiger wrote:

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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[*]

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Thus

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[*]

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[/list]

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

[list=*]
[*]

[/*]
[/list]


Bassaricyon neblina

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#5 2015-01-21 02:13:34

thedarktiger
Member
Registered: 2014-01-10
Posts: 91

Re: Infintite sieries

Thanks for the help guys!


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