Question

Question #bfe81

Answer 1

`(ln(x^2+1))/x^2=sum_(n=1)^oo(-1)^(n+1)/nx^(2n-2)=1-x^2/2+x^4/3-x^6/4...`

Explanation:

We know the following Maclaurin series for `ln(x+1)`:
`ln(x+1)=sum_(n=1)^oo(-1)^(n+1)/nx^n=x-x^2/2+x^3/3...`

We can find a series for `ln(x^2+1)` by replacing all the `x`'s with `x^2`:
`ln(x^2+1)=sum_(n=1)^oo(-1)^(n+1)/n(x^2)^n`

Now we can just divide by `x^2` to find the series we are looking for:
`(ln(x^2+1))/x^2=1/x^2sum_(n=1)^oo(-1)^(n+1)/nx^(2n)=`

`=sum_(n=1)^oo(-1)^(n+1)/n*x^(2n)/x^2=sum_(n=1)^oo(-1)^(n+1)/nx^(2n-2)=`

`=x^(2-2)-x^(2*2-2)/2+x^(3*2-2)/3-x^(4*2-2)/4...=`

`=1-x^2/2+x^4/3-x^6/4...`

which is the series we were looking for.