# The value of series $\sum_{n=1}^{\infty}\frac{n}{2^n}$ [duplicate]

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

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 $\left( \sum_{n=1}^\infty \frac{n}{2^n} \right) = 2$.

• why can we do this with an infinity sum, i this alowed? which rules did you used? Jan 1, 2016 at 10:56
• I just used linearity of the limit and index shifting, both of which are basic facts about convergent series. Jan 1, 2016 at 10:59
• often the binomial formulas are not basic facts for the students! Jan 1, 2016 at 11:01
• "Binomial"? What the...
– Did
Jan 1, 2016 at 11:03

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

• @DunVatta. You are very welcome ! Cheers and remember the trick (you will need to use it a lot of time). Jan 1, 2016 at 13:46

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 $x=\frac{1}{2}$

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.

• No: the complete series is easier.
– Did
Jan 1, 2016 at 10:31
• 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 Jan 1, 2016 at 10:33
• have a short look in a book for analysis Jan 1, 2016 at 10:34
• both authors haven't anythink said if such operations with an infinity sum are allowed this gives $-1$ and not for me Jan 1, 2016 at 10:36
• How easy is it to build it without a CAS ? Jan 1, 2016 at 15:28