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Hi Folks,

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

Given

[tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}[/tex]

I the sum into sub-sums

[tex]\sum_{n=0} ^{\infty} 2^{-n} + \sum_{n=0} ^{\infty} \frac{1}{2}^{n-1} [/tex]

taking [tex]2^{-n}[/tex]

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

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

I multiply with x on both sides and obtain

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

if I set x = 1 on both sides I get

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

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

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

Given

[tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}[/tex]

I the sum into sub-sums

[tex]\sum_{n=0} ^{\infty} 2^{-n} + \sum_{n=0} ^{\infty} \frac{1}{2}^{n-1} [/tex]

taking [tex]2^{-n}[/tex]

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

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

I multiply with x on both sides and obtain

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

if I set x = 1 on both sides I get

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

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

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