If $x=\dfrac {1}{1+y}, y>0$, prove that:

Question

If $x=\dfrac {1}{1+y}, y>0$, prove that: $x+x^2+x^3+.......\infty = \dfrac {1}{y}$

My Attempt: $$x=\dfrac {1}{1+y}$$ If $y>0$ then $x<1$ And, $x+x^2+x^3+......\infty$ is an infinite series.

Answer 1

Hint: $\;x=\dfrac {1}{1+y} \iff y = \dfrac{1}{x} - 1 \iff \dfrac{1}{y} = \dfrac{x}{1-x} = x \cdot \dfrac{1}{1-x}=x \cdot \left(1+x+x^2+ \ldots\right)\,$.

Answer 2

Guide:

Use the formula for geometric series,$$\sum_{i=1}^\infty x^i = \frac{x}{1-x}=\frac{1}{\frac1x-1}$$

Answer 3

It is the sum of a geometric serie $-1$ so it is ${1\over{1-{1\over {1+y}}}}-1$

$={1\over{{1+y-1}\over {1+y}}}-1={y+1\over y}-1={1\over y}$

Answer 4

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

Therefore $$\sum_{i=0}^n x^i=\frac{1-x^{n+1}}{1-x}$$ We know that $x<1$ and therefore $x^{n+1} \to_{n \to \infty} 0$ and therefore: $$\sum_{n=0}^\infty x^n= \frac{1}{1-x}$$ Multiplying by $x(= \frac{1}{1+y})$: $$\sum_{n=1}^\infty x^n= \frac{x}{1-x}=\frac{1}{y}$$

About the last equality, interrogate $x=\frac{1}{1+y}$ long enough and you will get $\frac{1}{y}=\frac{x}{1-x}$ :)

Answer 5

Notice that for any finite $n$ that.

$(1-x)(1 + x + x^2 + x^3 + ......... + x^n) = $

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

$1 - x^{n+1}$

So if $K_n(x) = 1 + x + x^2 + ..... + x^n$. And if $x \ne 1$, then

$K_n(x) = \frac {1 - x^{n+1}}{1-x} = \frac 1{1-x} - \frac {x^{n+1}}{1-x}$.

That's true always for any positive integer $n$.

If $-1 < x < 1$ then $\lim_{n\to \infty} x^{n+1} = 0$ so

If is $-1 < x < 1$ then $1 + x + x^2 + ..... = \lim_{\to \infty} K_n(x) = \frac 1 {1-x} - \lim_{n\to \infty}\frac {x^{n+1}}{1-x}= \frac 1{1-x}$.

Note: this is only true if $|x|< 1$. If $|x| \ge 1$ you can see the sum $1 + x + x^2 + ....$ either blows up, or fluctuates back and forth... or blows up while it is flucuating back and forth.

So....

if $y > 0$ then $0< x = \frac 1{1+y} < 1$

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

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

So it's a matter of showing $\frac 1{1-x} - 1 = \frac 1{1-x} - \frac {1-x}{1-x}=$

$\frac {x}{1 -x} = \frac 1{1+y}*\frac 1{1 - \frac 1{1+y}}$

$= \frac 1{1+y}*\frac {1+y}{(1+ y) - 1}= \frac 1y$.