# Math Help - Binomial series expansion

1. ## Binomial series expansion

Hey everyone, I'm stuck on the last part of this question, but it might require something from the previous parts, so I'm going to start from the beginning, including my working out which should hopefully be correct. Also, this is my first time typing in codes, so don't laugh if I make a mistake .

I have been asked to expand the following equation up to and including the term $y^3$, for |y|<1:

$(1-y)^{-2}$

This expands to:
$1+2y+3y^3+4y^3$

Hence, or otherwise, I had to show that:

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

I did that by substituting y as $\frac{x}{1+x}$

I then had to write it in the form $(a+x)^n$ where a and n are integers, which should be $(1+x)^2$.

This last part asks me to 'find the set of values of x for which the series (above) is convergent", and I haven't a clue what to do next. I'm really stuck and any help is appreciated .

2. Originally Posted by SuumEorum
Hey everyone, I'm stuck on the last part of this question, but it might require something from the previous parts, so I'm going to start from the beginning, including my working out which should hopefully be correct. Also, this is my first time typing in codes, so don't laugh if I make a mistake .

I have been asked to expand the following equation up to and including the term $y^3$, for |y|<1:

$(1-y)^{-2}$

This expands to:
$1+2y+3y^3+4y^3$

Hence, or otherwise, I had to show that:

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

I did that by substituting y as $\frac{x}{1+x}$

I then had to write it in the form $(a+x)^n$ where a and n are integers, which should be $(1+x)^2$.

This last part asks me to 'find the set of values of x for which the series (above) is convergent", and I haven't a clue what to do next. I'm really stuck and any help is appreciated .
If you are talking about $\sum_{n=1}^{\infty}\frac{nx^{n-1}}{(x+1)^{n-1}}$

THen applying the root test we would get

$\bigg|\frac{x}{x+1}\bigg|<1$

Which implies that

$\frac{x}{1+x}<1$

and $\frac{x}{1+x}>-1$

I think you can go from there

3. Oooh thanks for your help, I've never encountered the root test before but I've read all about it on Wiki now . But I'm still slightly confused, how come, after you've applied the root test you get $\bigg|\frac{x}{x+1}\bigg|<1$ but not $\bigg|\frac{nx}{x+1}\bigg|<1$?

4. Originally Posted by SuumEorum
Oooh thanks for your help, I've never encountered the root test before but I've read all about it on Wiki now . But I'm still slightly confused, how come, after you've applied the root test you get $\bigg|\frac{x}{x+1}\bigg|<1$ but not $\bigg|\frac{nx}{x+1}\bigg|<1$?
Note that $\lim_{n\rightarrow{+\infty}}\sqrt[n]{n}=1$ while $\lim_{n\rightarrow{+\infty}}\sqrt[n]{\frac{x^{n-1}}{(1+x)^{n-1}}}=\frac{x}{1+x}$ ( $x\geq{0}$)

5. Aaah I see it now, thanks , I can finally go to sleep .