Value of c if Summation n=2 to infinity of (1 + c)^(-n) = 2

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MarkSA

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What is the value of c, if the [Summation from n=2 to infinity of: (1 + c)^(-n)] is equal to 2?

I started by rewriting it as:

1/(1 + c)^n

Then took the index of the summation from n=2 to n=1 and it became:

1/(1 + c)^(n + 1)

I then rewrote it to get it in the geometric series form and it became:

1/[(1 + c)^2 * (1 + c)^(n-1)]

Then if a = (1 + c)^(-2) and r = (1 + c)^(-1), assuming -2 < c < 0 since that would mean -1 < r < 1

a/(1-r) = [1 + c^(-2)]/[1 - (1 + c)^(-1)]

After lots of messing with it I get a result of:

1/c(1 + c)

BTW is it proper to say that S(sub n) = 1/c(1 + c)? S(sub n) means the summation of the series, right?

Ok.. here I let 1/c(1 + c) = 2. I end up doing the quadratic on it and get c = [-1 +/- sqrt(3)]/2.

Here is my problem. The book says the right answer is c = [-1 + sqrt(3)]/2 . However, if I plug this into calculator I end up with 0.366. This is not in the range of the -2 < c < 0 that I had to assume above for the geometric series formula to work. How can this be so?
 
how did you get -2 < c < 0 ?

the common ratio for the geometric series is 1/(1+c), therefore -1 < 1/(1+c) < 1

so 1+c > 1 ... c > 0
or 1+c < -1 ... c < -2
 
I see:
1/(1 + c) < 1
1 + c > 1
c > 0

But also:
1/(1 + c) > -1
-(1 + c) < 1
-1 - c < 1
- c < 2
c > -2?

How is one to intrepret this? (this is something i've forgotten....)
 
MarkSA said:
I see:
1/(1 + c) < 1
1 + c > 1
c > 0

1/(1 + c) > -1
reciprocating both sides changes the sign of the inequality ...
1 + c < -1
c < -2
Click to expand...
 
1/(1 + c) > -1
reciprocating both sides changes the sign of the inequality ...
1 + c < -1
c < -2
Click to expand...

If I multiply both sides by (1 + c), I start by letting:
1/(1+c) > -1 (which is part of what I need in order to do the a/1-r rule..)
And get:
1 > -(1 + c)
Multiply both sides by negative, so switch inequality direction:
-1 < (1 + c)
-2 < c which equals c > -2

Can you tell which step I am going wrong at?
 
here's the screw up ... when you multiplied both sides by (1 + c), you assumed that the value of (1 + c) > 0.

go back to the original inequality ...

-1 < 1/(1 + c) < 1

the value of (1 + c) has to be either positive or negative ... it can't be zero.

case #1 ... assume (1 + c) > 0 ... multiply all terms by (1 + c)

-(1 + c) < 1 and 1 < 1 + c

solve both inequalities ...

1 + c > -1 and 1 < 1 + c
c > -2 and c > 0

... the intersection of these two solution sets is c > 0.

case #2 ... now assume (1 + c) < 0 ... multiply all terms by (1 + c)

-(1 + c) > 1 and 1 > 1 + c
1 + c < -1 and 1 > 1 + c
c < -2 and c < 0

what is the intersection of these two solution sets?
 
\(\displaystyle \sum\limits_{k = 2}^\infty {\left( {1 + c} \right)^{ - k} } = \frac{{\left( {1 + c} \right)^{ - 2} }}{{1 - \left( {1 + c} \right)^{ - 1} }} = 2\quad \Rightarrow \quad c = - \frac{1}{2} +\frac{{\sqrt 3 }}{3}\)

The other solution for c is out of range.
 
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