The value of series
$\sum_{n=1}^{\infty}\frac{n}{2^n}$
I try to write some terms,but of no use. Is there any general method to approach such questions.
Thanks
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2See this. – David Mitra Jan 01 '16 at 10:29
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This is one of the most asked questions about series: http://math.stackexchange.com/questions/tagged/sequences-and-series?sort=frequent&pageSize=50 – Hans Lundmark Jan 01 '16 at 10:30
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Notice that \begin{align*} \sum_{n=1}^\infty \frac{n}{2^n} &= \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^{n-1}} = \frac{1}{2} \sum_{n=0}^\infty \frac{n+1}{2^n} = \frac{1}{2} \left( \sum_{n=0}^\infty \frac{n}{2^n} + \sum_{n=0}^\infty \frac{1}{2^n} \right) \\ &= \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^n} + 1, \end{align*} thus $\left( \sum_{n=1}^\infty \frac{n}{2^n} \right) = 2$.
Jendrik Stelzner
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why can we do this with an infinity sum, i this alowed? which rules did you used? – Dr. Sonnhard Graubner Jan 01 '16 at 10:56
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2I just used linearity of the limit and index shifting, both of which are basic facts about convergent series. – Jendrik Stelzner Jan 01 '16 at 10:59
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often the binomial formulas are not basic facts for the students! – Dr. Sonnhard Graubner Jan 01 '16 at 11:01
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Hint
Consider $$S=\sum_{n=1}^{\infty}n x^n=x\sum_{n=1}^{\infty}n x^{n-1}=x \frac d {dx}\Big(\sum_{n=1}^{\infty} x^{n}\Big)$$
When you finish, replace $x$ by $\frac 12$.
Happy New Year
Claude Leibovici
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@DunVatta. You are very welcome ! Cheers and remember the trick (you will need to use it a lot of time). – Claude Leibovici Jan 01 '16 at 13:46
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Hint: $$\sum_{n=0}^{\infty }x^n=\frac{1}{1-x}$$ $$\sum_{n=1}^{\infty }nx^{n-1}=(\frac{1}{1-x})'$$ $$\sum_{n=1}^{\infty }nx^{n}=x(\frac{1}{1-x})'$$ plug $x=\frac{1}{2}$
E.H.E
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prove by induction that $$\sum_{i=1}^n\frac{i}{2^i}=-2\, \left( 1/2 \right) ^{n+1} \left( n+1 \right) -2\, \left( 1/2 \right) ^{n+1}+2 $$ and then calculate the limit.
Dr. Sonnhard Graubner
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yes, this is one way to find the sum for $n$ tends to infinity. Your statement is wrong this is the partial sum and with that we can prove if the infinity sum exists – Dr. Sonnhard Graubner Jan 01 '16 at 10:33
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both authors haven't anythink said if such operations with an infinity sum are allowed this gives $-1$ and not for me – Dr. Sonnhard Graubner Jan 01 '16 at 10:36
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the OP must also do something thank you for your hint the searched limit is in this form better to find – Dr. Sonnhard Graubner Jan 01 '16 at 11:02
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