Pseudo arithmetic or pseudo geometric?

$\begin{aligned} A & = \frac 12 + \frac 1{2^2} + \frac 1{2^3} + \frac 1{2^4} + \cdots \\ B & = \frac 02 + \frac 1{2^2} + \frac 2{2^3} + \frac 3{2^4} + \cdots \\ & = \frac 12\left(\frac 12 + \frac 2{2^2} + \frac 3{2^3} + \frac 4{2^4} + \cdots \right) \\ 2B & = \frac 12 + \frac 2{2^2} + \frac 3{2^3} + \frac 4{2^4} + \cdots \\ 2B - A & = 0 + \frac 1{2^2} + \frac 2{2^3} + \frac 3{2^4} + \cdots \\ 2B - A & = B \\ \implies B & = A \end{aligned}$

$\implies$ They are equal .

We note that $A$ is a geometric progression and it converges as $A = \displaystyle \sum_{n=1}^\infty \frac 1{2^n} = \frac 12 \sum_{n=0}^\infty \frac 1{2^n} = \frac 12 \left(\frac 1{1-\frac 12}\right) = 1$ . Therefore, the answer is valid.

Alternatively, we can split $B$ into a (countably) infinite number of infinite geometric summations, each of which sum to a term in $A$ :

$\begin{aligned} B &= \dfrac{1}{2^2} + \dfrac{2}{2^3} + \dfrac{3}{2^4} + \cdots \\ &= \dfrac{1}{2^2} + \left ( \dfrac{1}{2^3} + \dfrac{1}{2^3} \right ) + \left ( \dfrac{1}{2^4} + \dfrac{1}{2^4} + \dfrac{1}{2^4} \right ) + \cdots \\ &= \left ( \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4} + \cdots \right ) + \left ( \dfrac{1}{2^3} + \dfrac{1}{2^4} + \cdots \right ) + \left ( \dfrac{1}{2^4} + \cdots \right )+ \cdots \\ &= \dfrac{\frac{1}{2^2}}{1 - \frac{1}{2}} + \dfrac{\frac{1}{2^3}}{1 - \frac{1}{2}} + \dfrac{\frac{1}{2^4}}{1 - \frac{1}{2}} + \cdots \\ &= \dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots \\ &= A \end{aligned}$

(We can do this because the partial sums obviously converge uniformly to the overall series.) We conclude that $\boxed{A = B}.$

Since the first sum is geometrically increasing with common ratio $(r)$ $\dfrac{1}{2}$ which is evaluted as below. $\begin{aligned} & A = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +\cdots \\ & A = \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{2^n} = \dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}} = 1 \quad {\color{#3D99F6} |r|<1} \end{aligned}$ To evaluate the second sum we must note that Arithemetico-Geomertic Series . Rewriting the given sum as follows. $\begin{aligned} & B = \dfrac{1}{2}\left(0+\dfrac{1}{2}+ \dfrac{2}{4} + \dfrac{3}{8} + \cdots \right) \\& 2B = 0 +1.\dfrac{1}{2} +2.\dfrac{1}{4} + 3.\dfrac{1}{8} + \cdots \end{aligned}$ $\begin{aligned}& 2B = 0 + \dfrac{1}{2} + \dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+\cdots\\& \dfrac{2B}{2} =B = \quad 0+ \dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{16}+\dfrac{4}{32}+\cdots \end{aligned}$ Subtracting the above sums as follows. $\begin{aligned} & 2B- B = \left(\dfrac{1}{2}-0\right) + \left(\dfrac{2-1}{4}\right) +\left(\dfrac{3-2}{8}\right)+\left(\dfrac{4-3}{16}\right) \cdots \\& B = \dfrac{1}{2} + \dfrac{1}{4}+ \dfrac{1}{8}+\dfrac{1}{16}+ \cdots \\& B =\left(\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2} }\right) = 1\end{aligned}$ Shows that sums are equal ie $A =B$ .

$\begin{aligned}B & = \sum_{k=0}^\infty \frac k {2^{k+1}} = \sum_{k=0}^\infty \sum_{n = 1}^k \frac 1{2^{k+1}} \\ & = \sum_{n=1}^\infty \sum_{k = n}^\infty \frac 1{2^{k+1}} \\ & = \sum_{n=1}^\infty \frac 1{2^n} \sum_{k=0}^\infty \frac 1{2^{k+1}} \\ & = \sum_{n=1}^\infty \frac 1{2^n}\cdot 1 = A.\end{aligned}$ (And, of course, $A = B = 1$ .)

The trick lies in the switch from $\sum_{k=0}^\infty \sum_{n = 1}^k \dots$ to $\sum_{n=1}^\infty \sum_{k = n}^\infty \dots$ ; if you draw the range of addition in the $(k,n)$ -plane, it will be a triangle, which may be described as an infinite number of finite rows, or an infinite number of infinite columns.

First series is a geometric series $\sum_{n=1}^\infty x^n = \frac x{1-x} .$ The second one is $\sum_{n=0}^\infty nx^{n+1} = x^2 \sum_{n=0}^\infty nx^{n-1} = x^2 \left(\sum_{n=0}^\infty x^n\right)' = x^2 \left( \frac 1{1-x} \right)' = \frac{x^2}{(1-x)^2}$ . Evaluated at $x = 1/2$ both series give $1$ , so $\boxed {A=B}$ .

Many Mathematicians could see that a finite series of A 'gets closer and closer to 1 without reaching it', and so the infinite series tends to 1. Now to B which is a bit trickier. B can be written B = (1/4 + 1/8 + 1/16 + 1/32 + 1/64 + .....) + (1/8 + 1/16 + 1/32 + 1/64 + .......) + (1/16 + 1/32 + 1/64 + .....) + (1/32 + 1/64 + .....) + (1/64 + ...) + .... Note that the first part of this is one half of A. The 2nd part is 1/4 of A. The 3rd part is 1/8 of A etc. And A tended to 1, as explained in the first sentence. So B = (1/2 + 1/4 + 1/8 + .....) x 1. And the finite series of this has already been seen to 'get closer and closer to 1', so with an infinite number of terms the series tends to 1. So B = 1, the same as A. Regards, David

For me, I went with the approach of using the generating functions for the sequences for A and B (actually for 2A and 2B)

Let $\displaystyle S(x) = 1 + \frac {x}{2} + \frac {x^{2}}{2^{2}} + \frac {x^{3}}{2^{3}} + \ldots$

Notice that $S(1) = 2A$ .

Written with summation notation:
1 $\displaystyle S(x) = \sum_{i=0}^{\infty} \frac {x^{i}}{2^{i}}$

This sums to (sum of a geometric series with common ratio $\frac {x}{2}$ , when $-2 < x < 2$ ):
2 $\displaystyle S(x) = \frac {1}{1 - \frac {x}{2}}$

Therefore we get from 2 that $2A = S(1) = 2$ , or $A = 1$

Now taking the derivative of $S(x)$ wrt $x$ gives on one hand, from 1
$\displaystyle S'(x) = \sum_{i=0}^{\infty} \frac {i x^{ i - 1 }}{2^{i}}$
$\displaystyle S'(x) = 0 + \frac {1}{2} + \frac {2x}{2^{2}} + \frac {3x^{2}}{2^{3}} + \ldots$
So again, $S'(1) = 2B$ .

But we can also take the derivative another way, using the chain rule on 2 :
$\displaystyle S'(x) = - \frac {1}{(1 - \frac {x}{2})^{2}} (- \frac {1}{2})$
3 $\displaystyle S'(x) = \frac {\frac {1}{2}}{(1 - \frac {x}{2})^{2}}$

Therefore we get from 3 that $2B = S'(1) = 2$ , or $B = 1$

And so $A = B$

This is legitimate because the series are absolutely convergent.

A=1; 2(B-A)=-1+B; B-2A=-1; B=2A-1; B=1

Isn't this the (geometric) series you get when you solve last week's probability problem? Problem 1, 19 march

Taylor expend $\frac{1}{4(1-x)^2}$ and put x= $\frac{1}{2}$ in , you'll meet the second series.

I just matched values from B to values from A. Since the sequences are infinite you will be able to say A = B. For example, the first number in B, 0/2 has no value so I added 1/2x2 and 2/2x2x2 to get 1/4 + 1/4 which equals 1/2. Which is the first value for A. The second value in A is 1/4 which can be matched by the fourth value in B ( 3/16 ) plus part of the fifth value (2/16). You have 1/16 left over in B. That matches the 4th value in A. The third value in A can be matched by the 6th and 7th values in B. You can continue matching in this way because your denominators are the same so some parts will always match up. Because both series go on to infinity you will always be able to find a match for the values in A thus they are approaching the same limit.

B=A/2+A/4+A/8+.....=A

B= 1/2 2 +1/2 2 2 + 1/2 2 2 2.... +1/2 2 2 +1/2 2 2 2... . . . =1/2(1/2 +1/2 2 2 + 1/2 2 2 2 ...) +1/4(1/2+1/2 2 2+1/2 2 2*2....) +1/8(....) . . . =(1/2 +1/4 +1/8 +1/6..)A =A