1. ## Limits and integration

If I have shown that $\sum\limits_{k = 0}^\infty {{f_k}}$ is uniformly convergent on [0,1], where ${f_k}\left( x \right) = {\left( {1 - {x^2}} \right)^2}{x^k}$, how would I then go about showing that

$\int\limits_0^1 {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{{1 - x}}} {\rm{ dx}} = \sum\limits_{k = 1}^\infty {\frac{8}{{k\left( {k + 2} \right)\left( {k + 4} \right)}}}$?

All help appreciated!

2. Hello,
Originally Posted by converge2
If I have shown that $\sum\limits_{k = 0}^\infty {{f_k}}$ is uniformly convergent on [0,1], where ${f_k}\left( x \right) = {\left( {1 - {x^2}} \right)^2}{x^k}$, how would I then go about showing that

$\int\limits_0^1 {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{{1 - x}}} {\rm{ dx}} = \sum\limits_{k = 1}^\infty {\frac{8}{{k\left( {k + 2} \right)\left( {k + 4} \right)}}}$?

All help appreciated!
$\int_0^1 \frac{(1-x^2)^2}{1-x} ~dx=\int_0^1 (1-x^2)^2\sum_{k=0}^\infty x^k~dx=\int_0^1 \sum_{k=0}^\infty (1-x^2)^2x^k ~dx$

Since $\sum_{k=0}^\infty f_k(x)$ converges uniformly, we can invert the sum and the integral.

And hence $\int_0^1 \frac{(1-x^2)^2}{1-x} ~dx=\sum_{k=0}^\infty \int_0^1 (1-x^2)^2 x^k ~dx$

can you finish it ? expand $(1-x^2)^2$ and be careful with the case k=1.

3. Originally Posted by converge2
If I have shown that $\sum\limits_{k = 0}^\infty {{f_k}}$ is uniformly convergent on [0,1], where ${f_k}\left( x \right) = {\left( {1 - {x^2}} \right)^2}{x^k}$, how would I then go about showing that

$\int\limits_0^1 {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{{1 - x}}} {\rm{ dx}} = \sum\limits_{k = 1}^\infty {\frac{8}{{k\left( {k + 2} \right)\left( {k + 4} \right)}}}$?

All help appreciated!
How about expanding the integrand as a power series and the integrating term by term?

CB