limit of a sum: $\sum_{n} = 1+x+2x^2+3x^3+...+nx^n$

Question

I have a sum $$\sum_{n} = 1+x+2x^2+3x^3+...+nx^n$$ and asked to simplify it and given $|x|<1$ determine the limit as $n \to \infty$

My first impression is that this is a simple geometric series but I have a mental block as to what is the common ratio. I assume that it was simply $\sum_{n=0}^\infty nx^n$ but when I plug in values to test it the first terms don't seem intuitive. But following from this would the common ratio be x.

https://en.m.wikipedia.org/wiki/Arithmetico–geometric_sequence#Infinite_series — lab bhattacharjee, Apr 17 '18 at 14:07
https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-inftyn1xn — Bill O'Haran, Apr 17 '18 at 14:07

PM. · Accepted Answer · 2018-04-18T15:54:33.870

$$\text{Let }\; S = 1+x+2x^2+3x^3+...+nx^n. $$

$$ \text{then }\;xS=x+x^2+2x^3+3x^4+...+nx^{n+1} $$

$$\text{Subtract: }\: S(1-x)=1+x^2+x^3+\ldots+x^n-nx^{n+1} $$

$$\text{Note that }\; x^2+x^3+\ldots+x^n $$ is a geometric series, with first term $x^2$, common ratio $x$ and $n-1$ terms.

You can compute that geometric series and simplify the overall expression for $S$.

For $n \rightarrow \infty $ then $x^n \rightarrow 0$ when $|x| < 1$.

score 1 · Answer 2 · answered Apr 17 '18 at 18:41

Note that $$ 1+x+x^2+x^3+.... = \frac {1}{1-x} $$

Differentiate to get

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

term by term differentiation implies

$$ \frac {d}{dx} (1+x+x^2+x^3+....)= 1+2x+3x^2+.... $$ Therefore

$$ 1+2x+3x^2+.... = \frac {1}{(1-x)^2} $$

Multiply by x

$$ x+2x^2+3x^3+...=\frac {x}{(1-x)^2}$$

add $1$ to bothe sides

$$ 1+x+2x^2+3x^3+... =1+ \frac {x}{(1-x)^2}$$

Mike · Answer 3 · 2018-04-17T14:21:22.050

0

On the interval $(-1,1)$, the infinite sum $1+x+x^2+x^3 \ldots$ $=\frac{1}{1-x}$, which we will write as $y(x)$.

The sum you wrote is $1+ x \left(\frac{dy(x)}{dx}\right)$.

edited Apr 17 '18 at 14:21

answered Apr 17 '18 at 14:15

Mike

20,434

score 0 · Answer 4 · answered Apr 17 '18 at 18:56

$$ 1+x+2x^2+3x^3+... = $$

$$(1+x +x^2 +x^3+...)+ (x^2 + x^3 +x^4+...)+ ( x^3 + x^4 +x^5 +...)+...=$$

$$ ( \frac {1}{1-x}) +(\frac {x^2}{1-x})+ (\frac {x^3}{1-x})+...=$$ $$ ( \frac {1}{1-x})(1+x^2 +x^3 +...)=$$

$$ ( \frac {1}{1-x})(1+\frac {x^2}{1-x}) =$$

$$ \frac {1-x+x^2}{(1-x)^2}$$

score 0 · Answer 5 · answered Apr 18 '18 at 00:52

I have already shown in another post that:

$S=1+2x+3x^2+4x^3+ . . .nx^{n-1}=\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}$

Multiplying bth sides by x we get:

$x.S=x+2x^2+3x^3+ . . .nx^n ⇒x.S+1=1+x+2x^2+3x^3+ . . .nx^n =s $

so the answer is:

$s=x.S+1=x.\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}+1=\frac{{nx^{n+2}-(n+1)x^{n+1}+x+(x-1)^2}}{(x-1)^2}$

limit of a sum: $\sum_{n} = 1+x+2x^2+3x^3+...+nx^n$

5 Answers5