Smart demonstration to the formula $ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$

Question

Someone could give me a smart and simple solution to show the folowing identity?

$$ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$$

Answer 1

$$\sum_{n=1}^N \dfrac{n}{2^n}$$ can be written as \begin{matrix} \dfrac12 & + \dfrac1{2^2} & + \dfrac1{2^3} & + \dfrac1{2^4} & + \cdots & + \dfrac1{2^{N-1}} & + \dfrac1{2^N}\\ & + \dfrac1{2^2} & + \dfrac1{2^3} & + \dfrac1{2^4} & + \cdots & + \dfrac1{2^{N-1}} & + \dfrac1{2^N}\\ & & + \dfrac1{2^3} & + \dfrac1{2^4} & + \cdots & + \dfrac1{2^{N-1}} & + \dfrac1{2^N}\\ & & & + \vdots & + \vdots & + \vdots & + \vdots\\ & & & & & & + \dfrac1{2^N}\\ \end{matrix} Now sum them row wise to get $$\left(1 - \dfrac1{2^N} \right) + \dfrac12\left(1 - \dfrac1{2^{N-1}} \right) + \dfrac1{2^2}\left(1 - \dfrac1{2^{N-2}} \right) + \cdots + \dfrac1{2^{N-1}}\left(1 - \dfrac1{2} \right)\\ = \left(1 + \dfrac12 + \dfrac1{2^2} + \cdots + \dfrac1{2^{N-1}} \right) - \left(\dfrac1{2^N} + \dfrac1{2^N} + \dfrac1{2^N} + \cdots + \dfrac1{2^N} \right)\\ = 2 - \dfrac1{2^{N-1}} - \dfrac{N}{2^N}$$ Essentially, we are doing the following $$\sum_{n=1}^N \dfrac{n}{2^n} = \sum_{n=1}^N \sum_{k=1}^n \dfrac1{2^n} = \sum_{k=1}^N \sum_{n=k}^{N} \dfrac1{2^n} = \sum_{k=1}^N \dfrac1{2^{k-1}}\left(1 - \dfrac1{2^{N+1-k}}\right) = \sum_{k=1}^N \left(\dfrac1{2^{k-1}} - \dfrac1{2^{N}} \right)\\ = 2 - \dfrac1{2^{N-1}} - \dfrac{N}{2^N}$$

Answer 2

See the following answer: How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$.

We can use a similar technique as that used to sum the geometric series $\sum_{n=1}^k r^n.$ Let $$S_{m}=\sum_{n=1}^{m}nr^{n},$$ and consider $$S_{m}-rS_{m}=-mr^{m+1}+\sum_{n=1}^{m}r^{n}.$$ Using our known formula for the geometric series, we find that $$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$
This equality holds for any $r\neq 1$, so inserting $r=\frac{1}{2}$ we have that $$\sum_{n=1}^N \frac{n}{2^n}=\frac{2^N+1-N+2}{2^N}$$

Answer 3

$$\eqalign{\sum_{n=1}^N n r^n &= r \dfrac{d}{dr} \sum_{n=0}^n r^N \cr &= r \dfrac{d}{dr} \dfrac{1 - r^{N+1}}{1-r}\cr &= r \dfrac{-(N+1) r^N (1-r) + (1 - r^{N+1})}{(1 - r)^2}\cr}$$

Answer 4

Proceed by induction.

Base case is true with $N=1$.

Inductive (I replaced the actual sum with $\sum$): $$ \frac{N+1}{2^{N+1}} + \sum^N = \sum^{N+1} $$ $$ \frac{N+1}{2^{N+1}} + \frac{-N+2^{N+1}-2}{2^N} = \frac{-(N+1) + 2^{N+2}-2}{2^{N+1}} $$ $$ \frac{N + 1 - 2N + 2^{N+2} - 4}{2^{N+1}} = \frac{-N-3+2^{N+2}}{2^{N+1}} $$ $$ \frac{- N - 3 + 2^{N+2}}{2^{N+1}} = \frac{-N-3+2^{N+2}}{2^{N+1}} $$

Which is true, so the summation is true. This is a good general way for proving summation formulas.

Answer 5

This another approach shown as a way to use the Online Encyclopedia of Integer Sequences database to help construct a solution. If you take the following from the summation formula $$n=(1)=\frac{1}{2^1}=\frac{1}{2^1}$$ $$n=(2)=\frac{1}{2^1}+\frac{2}{2^2}=\frac{4}{2^2}$$ $$n=(3)=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}=\frac{11}{2^3}$$ $$n=(4)=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}=\frac{26}{2^4}$$ Note the sequence in the numerator. Go to OEIS and type in the sequence $1,4,11,26$ This will take you to sequence A000295 This sequence gives the formula $$a(n)=2^n - n - 1$$ However this sequence and formula happen to have a different offset (the subscript of the first term in the sequence) than the formula given by the OP. [A000295] is the following sequence $0,1,4,11,26...$ and we want $1,4,11,26...$. Taking the difference between these two sequences gives another sequence $1,3,7,15,31,...$. Enter this sequence in [OEIS] and it will take you to A000225. You will see that the formula for this sequence is $$2^n-1$$ Adding the formulas for [A000295] $2^n-n-1$ plus [A000225] $2^n-1$ give the desired formula for the numerator of $$2^{n+1}-n-2$$ It is obvious that the sum of the sequence 0,1,4,11,26 and 1,3,7,15,31 gives the desired sequence of 1,4,11,26,… with offset 1. To develop the formula yourself note that the page for [A00295] also gives the generating function for the series of $$\frac{x^2}{(1-2x)(1-x)^2}$$ from which the Binet formula can be derived using partial fractions as follows $$\frac{x^2}{(1-2x)(1-x)^2}=\frac{A}{(1-2x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^2}$$

We end up with $$A(1-2x+x^2)+B(1-3x+2x^2)+C(1-2x)$$ \begin{bmatrix} 1 & 1 & 1 & 0 \\[0.3em] -2 & -3 & -2&0 \\[0.3em] 1&2&0&1 \end{bmatrix} Solving the matrix we get $A=1, B=0, C=-1$ $$\eqalign{\sum_{n=0}^N 2^n x^n}+0\eqalign{\sum_{n=0}^N n x^n}-\eqalign{\sum_{n=0}^N (n+1) x^n}$$ yielding $2^{n+1}-n-1$ Or you can develop the generating function from the recurrence relation for this sequence. The same thing can be done with the series A00225 The page for [A00225] also gives the generating function for the series of $$\frac{x} {(1-2x)(1-x)}$$ from which the Binet formula can be derived using partial fractions as follows $$\frac{x}{(1-2x)(1-x)}=\frac{A}{(1-2x)}+\frac{B}{(1-x)}$$

We end up with $$A(1-x)+B(1-2x)$$ \begin{bmatrix} 1 & 1 & 0 \\[0.3em] -1 & -2 & 1 \end{bmatrix} Solving the matrix we get $A=1, B=-1$ $$\eqalign{\sum_{n=0}^N 2^n x^n}-\eqalign{\sum_{n=0}^N 1^n x^n}$$ yielding $2^n-1$. So $$\frac{(2^n-n-1)+(2^n-1)}{2^n}=\frac{2^{n+1}-n-2}{2^n}$$