# Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for ƒ (perhaps more than once). Give the interval of convergence for the resulting series. g(x)=-frac{1}{(1+x)^2}text{ using }f(x)=frac{1}{1+x}

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for ƒ (perhaps more than once). Give the interval of convergence for the resulting series.
$$g(x)=-\frac{1}{(1+x)^2}\text{ using }f(x)=\frac{1}{1+x}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Viktor Wiley

Given the series
$$f(x)=\frac{1}{1+x}$$
We have to find the power series of g(x) by differentiating or integrating f(x). Also we have to find the interval of convergence.
Formula used:
1.$$\frac{1}{1+x}=\sum_{n=0}^\infty x^n$$
2.$$\frac{d}{dx}(x^n)=nx^{n-1}$$
3. If $$\sum_{n=1}^\infty a_n(x-c)^n$$ is a power series with radius of convergence R, then the interval of convergence is $$c-R$$
4.The series $$\sum a_nx^n$$ is convergent then $$\lim_{n\rightarrow\infty}|\frac{a_n}{a_{n+1}}|>1$$
Now, Power series by differentiating f,
$$\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum_{n=0}^\infty(-x)^n=\sum_{n=0}^\infty(-1)^nx^n$$
$$\frac{1}{1+x}=\sum_{n=0}^\infty(-1)^nx^n$$
diff. both sides,
$$\frac{d}{dx}(\frac{1}{1+x})=\frac{d}{dx}(\sum_{n=0}^\infty(-1)^nx^n)$$
$$\frac{-1}{(1+x)^2}=\sum_{n=0}^\infty(-1)^nnx^{n-1}$$
$$\frac{-1}{(1+x)^2}=0+\sum_{n=1}^\infty(-1)^nnx^{n-1}$$
$$\frac{-1}{(1+x)^2}=\sum_{n=1}^\infty(-1)^nnx^{n-1}$$
Now to find the interval of convergence:
We have
$$a_n=(-1)^nnx^{n-1}$$
Now, $$\lim_{n\rightarrow\infty}|\frac{a_n}{a_{n+1}}|=\lim_{n\rightarrow\infty}|\frac{(-1)^nnx^{n-1}}{(-1)^{n+1}(n+1)x^n}|=\lim_{n\rightarrow\infty}|\frac{n}{-1(n+1)x}|=\lim_{n\rightarrow\infty}|\frac{1}{(1+\frac1n)x}|=|\frac1x|$$
now $$\lim_{n\rightarrow\infty}|\frac{a_n}{a_{n+1}}|=|\frac1x|>1$$
$$\frac{1}{|x|}>1$$
$$|x|<1\ i.e.\ -1$$
Now check at boundary points of the interval:
At $$x=1,\ \sum_{n=1}^\infty(-1)^nn(1)^{n-1}=\sum_{n=1}^\infty(-1)^nn$$
does not converge.
At $$x=-1,\sum_{n=1}^\infty(-1)^nn(-1)^{n-1}=\sum_{n=1}^\infty(-1)^{2n-1}n$$
does not converge.
Hence the interval of convergence is (-1, 1).

###### Not exactly what you’re looking for?
content_user

Answer is given below (on video)