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JD96

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Hello,

I want to prove that the taylor expansion of [itex]f(x)={\frac{1}{\sqrt{1-x}}}[/itex] converges to ƒ for -1<x<1. If I didn't make a mistake the maclaurin series should look like this:

[itex]Tf(x;0)=1+\sum_{n=1}^\infty{\frac{(2n)!}{(2^n n!)^2}}x^n[/itex]

My attempt is to use the lagrange error bound, which is given by [itex]E(x)\leq M\frac{|x|^{n+1}}{(n+1)!}[/itex], where M is the upper bound for [itex]|f^{(n+1)}(x)|[/itex] for the interval [0,x] and show it tends to zero for [itex]\lim_{n\rightarrow \infty}[/itex].

My problem is the following: For functions like e

[itex]\lim_{n\rightarrow \infty}(\frac{(2n)!}{4^n n!}(1-x)^{-\frac{2n+1}{2}}\frac{|x|^{n+1}}{(n+1)!})[/itex]=0

I have no clue how to solve such a limit and would appreciate any help! Of course I don't expect anyone to work it out, but maybe someone has an advice as how to tackle complicated limits like this one... .

Thanks in advance!

I want to prove that the taylor expansion of [itex]f(x)={\frac{1}{\sqrt{1-x}}}[/itex] converges to ƒ for -1<x<1. If I didn't make a mistake the maclaurin series should look like this:

[itex]Tf(x;0)=1+\sum_{n=1}^\infty{\frac{(2n)!}{(2^n n!)^2}}x^n[/itex]

My attempt is to use the lagrange error bound, which is given by [itex]E(x)\leq M\frac{|x|^{n+1}}{(n+1)!}[/itex], where M is the upper bound for [itex]|f^{(n+1)}(x)|[/itex] for the interval [0,x] and show it tends to zero for [itex]\lim_{n\rightarrow \infty}[/itex].

My problem is the following: For functions like e

^{x}or sin(x) M is independent of n and thus the limit is easy to compute, but here M is given by [itex]f^{(n+1)}(x)=\frac{(2n)!}{4^n n!}(1-x)^{-\frac{2n+1}{2}}[/itex] (which I conclude from graphing the function). So I need to prove that the following holds true (for -1<x<1):[itex]\lim_{n\rightarrow \infty}(\frac{(2n)!}{4^n n!}(1-x)^{-\frac{2n+1}{2}}\frac{|x|^{n+1}}{(n+1)!})[/itex]=0

I have no clue how to solve such a limit and would appreciate any help! Of course I don't expect anyone to work it out, but maybe someone has an advice as how to tackle complicated limits like this one... .

Thanks in advance!

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