- #1
Mathman23
- 254
- 0
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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