Final Review problems

Radius

for $f(x)=\sum k^2{x}^2$
$R={\lim }_k\sup \sqrt[k]{|k^2|}⟹R=1$ because

1. $k^{\frac{2}{k}}\ge 1⟹R\ge 1$
2. $R\le 1$ because
   $(1+\epsilon {)}^k\ge (1+\epsilon k)⟹(\exists k_0)(\forall k\ge k_0)(1+\epsilon {)}^k\ge (1+k\epsilon )\ge k^2$
   $\square$

Riemann integral

Calculate integral of ${\lim }_{a⟶1}\sum cos(a^{\frac{j}{2}})a^j(1-a)$
so mesh is like: $x_0=1,x_1=a,x_j=a^j,x_N=0$
$R_{\infty }(f)={\sum }_{j=0}^{\infty }f(a^j)(x_j-x_{j+1})$ where $x_j-x_{j+1}=a^j(1-a)$
take $x_j=a^j={\xi }_j$
so

$$R_{\infty }(f,\xi )={\int }_0^{1}f(t)dt={\int }_0^1\cos (\sqrt{x})dx$$

$\square$