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).