Is the Sum of a Geometric Series Always Equal to 2?

[Mathman23]

Hi Folks,

I have this here geometric series which I'm supposed to find the sum of:

Given

\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}

I the sum into sub-sums

\sum_{n=0} ^{\infty} 2^{-n} + \sum_{n=0} ^{\infty} \frac{1}{2}^{n-1}

taking 2^{-n}

Since x^n converges towards 1/1+x therefore I differentiate on both sides

1/(1+x)^2 = \sum_{n=0} ^{\infty} n \cdot 2^{-n} x^{n-1}

I multiply with x on both sides and obtain

x/(1+x)^2 = \sum_{n=0} ^{\infty} n * 2^{-n} x^n

if I set x = 1 on both sides I get

(1/4) ? = \sum_{n=1} ^{\infty} n*2^{-n} = 2

My teacher says that the expression has to give 2 on the left side, and not (1/4).

What am I doing wrong? Any surgestions?

Best Regards
Fred

 

[AKG]

Your first line is wrong. When you break your sum up into two sums, you don't do it right. Also, when you say xⁿ converges towards 1/1+x, I suppose you mean \sum x^n converges to 1/1+x. However, this is wrong, it converges to 1/1-x. After that, I have no idea what you're doing. I don't know where you're getting the "=" sign from. Sorry, I think just about everything in your post makes no sense.

 

[HallsofIvy]

\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}

is not a geometric series!

 

[Curious3141]

Hi Folks,

I have this here geometric series which I'm supposed to find the sum of:

Given

\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}

I the sum into sub-sums

\sum_{n=0} ^{\infty} 2^{-n} + \sum_{n=0} ^{\infty} \frac{1}{2}^{n-1}

taking 2^{-n}

Since x^n converges towards 1/1+x therefore I differentiate on both sides

1/(1+x)^2 = \sum_{n=0} ^{\infty} n \cdot 2^{-n} x^{n-1}

I multiply with x on both sides and obtain

x/(1+x)^2 = \sum_{n=0} ^{\infty} n * 2^{-n} x^n

if I set x = 1 on both sides I get

(1/4) ? = \sum_{n=1} ^{\infty} n*2^{-n} = 2

My teacher says that the expression has to give 2 on the left side, and not (1/4).

What am I doing wrong? Any surgestions?

Best Regards
Fred

Firstly, you separated the sum wrongly. Here's the correct version :

\sum_{n=0} ^{\infty} \frac{2n+1}{2^n} = \sum_{n=0} ^{\infty}(n.2^{1-n})+\sum_{n=0} ^{\infty}(2^{-n})

Hint : For the first part-sum on the RHS, try dividing by 2 then subtracting away from the original part-sum.

 

[Curious3141]

\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}

is not a geometric series!

Actually, it works out to be one !