Juxtaposing sum

Let us assume that the first series is equal to $2$ .

$1 + x + x^2 + x^3 + \cdots=2$

Now, let us try try to convert the second series into known values from the first.

Let $x + 2x^2 + 3x^3 + 4x^4 + \cdots=S$

$S=(x+x^2+x^3\cdots)+(x^2+x^3+x^4\cdots)+(x^3+x^4+x^5\cdots)+\cdots$

From the first series we know that $1+x+x^2+x^3\cdots=2$

$\Rightarrow x+x^2+x^3\cdots=1$ . Now putting this in $S$ we get

$\Rightarrow S=(1)+(1x)+(1x^2)+\cdots$

$\Rightarrow S= 1 + x + x^2 + x^3 + \cdots$

So we find that the value of the first and second series is same.

So the answer is $\boxed{2}$

Relevant wiki: Arithmetic-Geometric Progression

Let $\ S = 1 + x + x^2 + x^3 + ....$ and $\ T = x + 2x^2 + 3x^3 + 4x^4 + ....$ . We know that as a geometric series, $S$ will converge for $|x| < 1$ . Using the ratio test on $\ T$ ,

$\lim_{n \to \infty} \Bigg\lvert \dfrac{a_{n + 1}}{a_n} \Bigg\rvert = \lim_{n \to \infty} \Bigg\lvert \dfrac{(n + 1)\ x^{n + 1}}{n\ x^n} \Bigg\rvert = |x| \cdot \lim_{n \to \infty} \Bigg\lvert \dfrac{n + 1}{n} \Bigg\rvert = |x|$

so $\ T$ will also converge for $\ |x| < 1$ , and since $\ T$ will clearly not converge for $\ x = \pm 1, \ S$ and $\ T$ will converge for the same values of $x$ .

We know that for $|x| < 1, \ S = \dfrac{1}{1 - x}$ . We can develop a similar formula for $\ T$ .

$\begin{aligned} xS &= \ x + x^2 + x^3 + x^4 + .... \\ \\ T - xS &= \ x^2 + 2x^3 + 3x^4 + 4x^5 + .... = \ xT \\ \\ T - xT &= \ xS \\ \\ T(1 - x) &= \ \dfrac{x}{1 - x} \\ \\ T &= \ \dfrac{x}{(1 - x)^2} \end{aligned}$

Now we are ready to prove that $\ S = 2 \iff T = 2$ .

$\begin{aligned} S = 2 &\iff \dfrac{1}{1 - x} = 2 \\ \\ \\ &\iff 1 = 2 - 2x \\ \\ &\iff x = \dfrac{1}{2} \\ \\ \\ \\ \\ T = 2 &\iff \dfrac{x}{(1 - x)^2} = 2 \\ \\ \\ &\iff x = 2 (1 - x)^2 \\ \\ &\iff 2x^2 - 5x +2 = 0 \\ \\ &\iff (2x - 1)(x - 2) = 0 \qquad \small \text{[But we can disregard the solution } x = 2 \text{ as } |x| < 1] \\ \\ &\iff x = \dfrac{1}{2} \end{aligned}$

Since $\ S = 2$ and $\ T = 2$ are both equivalent to $\ x = \dfrac{1}{2}$ , it must be true that $\ S = 2 \iff T = 2$

Relevant wiki: Arithmetic-Geometric Progression

Let us call the first series A. Therefore $A=1+x+x^2+x^3+\ldots$

Let us call the second series B. Therefore $B=x+2x^2+3x^3+\ldots$

Intuitively, I found a nice equation relating the 2 series. That is $A^2-A=B$ (Proof is below)

$A^2=1+2x+3x^2+4x^3+\ldots$

Subtracting A from $A^2$ ,

$A^2-A=(1+2x+3x^2+4x^3+\ldots)-(1+x+x^2+x^3+x^4+\ldots)$

$A^2-A=x+2x^2+3x^3+4x^4+\ldots=B$

Therefore $A^2-A=B$

Using this equation, if we put the value of A=2, the value of $B=4-2=2$

If we put the value of B=2, the value of $A=2$

Therefore the correct answer is $\boxed{2}$

Suppose that $\sum_{n=0}^{\infty} x^n= 1+x^2 +x^3 +\cdots = \dfrac{1}{1-x} = 2 \phantom{ccc} {\color{#3D99F6} |x| <1 }$ Solving we get $x=\dfrac{1}{2}$ Then plugging back to the next series we find it as arithmatics geometric series as follows $S_2 = \dfrac{1}{2} +\dfrac{2}{4} + \dfrac{3}{8}+\cdots = \large{\boxed{2}}\qquad {\color{#3D99F6}\text{Note}}$

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Note:

$S_2 = \dfrac{1}{2} +\dfrac{2}{4} + \dfrac{3}{8} +\cdots = \dfrac{1}{2} +\dfrac{1+1}{4} +\dfrac{1+1+1}{8} +\cdots \\ S_2 = \left(\dfrac{1}{2} +\dfrac{1}{4} +\dfrac{1}{8}+\cdots\right) +\left(\dfrac{1}{4} +\dfrac{1}{8} +\dfrac{1}{16}+\cdots \right ) + \left(\dfrac{1}{8} +\cdots\right) +\cdots \\ S_2 = 1 + \dfrac{1}{2} + \dfrac{1}{8} +\dfrac{1}{16} + \cdots = \dfrac{1}{1-\frac{1}{2}} = \boxed{2}$

For $x<1$ ,

$y=1+x+x^2+x^3+x^4+\cdots=\dfrac{1}{1-x}=2,$

which yields $x=1/2$ . Differentiating both sides,

$\dfrac{dy}{dx}=1+2x+3x^2+4x^3+\cdots=\dfrac{1}{(1-x)^2}.$

Finally, multiply both sides by $x$ and plug $x$ in to get the result.

$x\dfrac{dy}{dx}=x+2x^2+3x^3+4x^4+\cdots=\dfrac{x}{(1-x)^2}=2$

Let us suppose the first series sums to 2. Then we have: $1+x^2+x^3+... = \frac{1}{1-x} = 2 \implies x = \frac{1}{2}$ .

Now consider the second series. Notice how it may be written as x times the derivative of the first series.

$x+2x^2+3x^3+4x^4+... = x \frac{d}{dx}\big( \frac{1}{1-x} \big) = \frac{x}{(1-x)^2} = \frac{1/2}{(1-1/2)^2} = 2$

Now suppose the second series sums to 2. By the same manipulation above, this leads to a quadratic equation with solutions $x = 2$ or $x = 1/2$ . We can dismiss the first solution because it leads to a divergent sum when substituted into the first series. The second solution yields a sum of $\boxed{2}$ when substituted into the first sum. Therefore, the first series sums to 2 if and only if the second series sums to 2.

Let $S_{1}$ be the first series, and $S_{2}$ the second series. $S_{1}$ is a geometric series and converges only if abs(x) less than 1. One find easily that $S_{2}=S_{1}(S_{1}-1)$ by making a table and summing in diagonal (the first row is $S_{1}$ , the first column is $(S_{1}-1)$ : the product starts from row2col2)

Hence if $S_{1}=2$ , then $S_{2}=2$ for $x=\frac{1}{2}$ . But the reverse problem: if $S_{2}=2$ , from $S_{2}=S_{1}(S_{1}-1)$ one obtains two values for $S_{1}$ : $S_{1}=2$ for $x=\frac{1}{2}$ or $S_{1}=-1$ for $x=2$ : but the second solution is impossible because the series would not converge. Sorry, I do not know how to write in LaTex

The first series is a decreasing geometric series and if it's sum is 2 then using the decreasing geometric series sum formula we get a common ratio of x=1/2. But you could also look at the first series as the Maclaurin Series for 1/(1-x). You can turn it into the Maclaurin Series for the second series by taking the first derivative and multiplying by x. The second series is then x/(1-x)^2. Simply let x=1/2 to yield 2.

I could solve this question anyway but I have a doubt.

Let's say $1 + x + x^2 + x^3 ..... = 2$

Therefore, $x + x^2 + x^3 ..... = 1$ ...... $(i)$

Let's take the second expression and write it as $x(1 + 2x + 3x^2 + 4x^3 .......)$

Further write it as $x(d/dx(x) + d/dx(x^2) + d/dx(x^3).....)$

= $x d/dx(x + x^2 + x^3 + x^4 ....... )$

From $(i)$ , we get $x d/dx (1)$

$= 0$ [as $d/dx (1) = 0$ ]

Where is the flaw in this procedure?

Let $|x| < 1$ .

$\sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}$

and,

$\sum_{n = 0}^{\infty} x^n = \dfrac{1}{1 - x}$

$\dfrac{1}{1 - x} = 2 \implies x = \dfrac{1}{2} \implies \dfrac{x}{(1 - x)^2} = \boxed{2}$ .

Now if we let $\dfrac{x}{(1 - x)^2} = 2 \implies 2x^2 - 5x + 2 = 0 \implies x = 2,\dfrac{1}{2}$ since $|x| < 1$ we choose $x = \dfrac{1}{2} \implies \dfrac{1}{1 - x} = \boxed{2}$ .

Let y=1+x+x^x+..... Now, y=1/(1-x) y=2, therefore x=1/2 derivative of y w.r.t x ={1/(1-x)}^2=1+2 x+3 x^2+......=(x+2x^2+3x^3+...)/x Hence, x+2x^2+3x^3+....=x*{1/(1-x)^2}=2

Let $A = 1 + x + x^2 + x^3 + \dots$ and $B = x + 2x^2 + 3x^3 + 4x^4 + \dots$ .

$A$ is a geometric series, which means that $A$ is convergent if and only if $|x| < 1$ , in which case one can write:

$\boxed{A = \displaystyle\frac{1}{1-x}}.$

Let's now write $B$ as a function of $x$ and $A$ :

$\begin{aligned} B &&=&& x && +&& 2x^2 && +&& 3x^3&& + && \dots \\ xB && =&& && &&x^2&& +&& 2x^3 &&+ &&\dots \\ B\left(1-x\right) &&=&& x &&+&& x^2 &&+&& x^3 &&+&& \dots \\ B\left(1-x\right) &&=&& xA && && && && && \end{aligned}$

Replacing $A$ with its expression: $\boxed{B = \displaystyle\frac{x}{\left(1-x\right)^2}}$

From a given value of either $A$ or $B$ , it is now straightforward to calculate a value for $x$ and then compute a value for the remaining series. Let's conclude by splitting the problem in two cases:

• Suppose $A = 2$ , then $x = 0.5$ and it follows that $B = 2$ .
• Suppose $B = 2$ , then solving the quadratic equation: $2x^2-5x+2=0$ , $x \in \{0.5; 2\}$ . But 2 is not allowed for $x$ as it would cause both series to blow up. So, $x = 0.5$ and it follows that $A = 2$ .

To sum up, either series being equal to 2 implies that the other series equals 2 as well.

for the first series it is easily to get S1=1/1-x then use it it to solve the sum for the second series . first time x on both side to xS2 = x^2 +2x^3+3x^4 ....... then get the xS2 = S2 -S1 +1 and from the equation above solve S2 = x/(1-x)^2 so if S1=2 then S2=2 and if S2=2 then S1 = 2 or -1 and since they don't have the answer of -1 thus the answer is 2

$1+x+x^2+x^3... = \sum_{n=0}^{\infty} x^n = \boxed{\dfrac{1}{1-x}}$ $x+2x^2+3x^3+4x^4... = \sum_{n=0}^{\infty} (n+1)x^{n+1} = \sum_{n=0}^{\infty} (n+2)x^{n+1} - \sum_{n=0}^{\infty} x^{n+1}$ $f(x) = \sum_{n=0}^{\infty} (n+2)x^{n+1}$ $\int f(x) = \sum_{n=0}^{\infty} x^{n+2} = \dfrac {x^2}{1-x}$ $f(x) = \dfrac {2x-x^2}{(1-x)^2}$ $\sum_{n=0}^{\infty} (n+1)x^{n+1} = \dfrac {2x-x^2}{(1-x)^2} - \dfrac{x}{1-x}= \boxed{\dfrac{x}{(1-x)^2}}$

If the first one is two, $x = 0.5$ , therefore the second one is also two.

If the second one is two, $x=0.5$ also. The other solution $x=2$ is outside the radius of convergence of the second series. Therefore, the first one is also two.