The real numbers
![a_0,a_1,a_2,\ldots](/media/m/b/0/4/b046b0c22cd770021f186732b35e6660.png)
satisfy
![1=a_0\le a_1\le a_2\le\ldots. b_1,b_2,b_3,\ldots](/media/m/6/9/8/698c06646e386db0e06f720bac135d44.png)
are defined by
![b_n=\sum_{k=1}^n{1-{a_{k-1}\over a_k}\over\sqrt a_k}](/media/m/7/b/5/7b5c3560a5725534b13fa7b83fafd0f4.png)
.
a.) Prove that
![0\le b_n<2](/media/m/c/4/b/c4b47c0829ca8be227af25012950b87f.png)
.
b.) Given
![c](/media/m/e/a/3/ea344283b6fa26e4a02989dd1fb52a51.png)
satisfying
![0\le c<2](/media/m/4/8/c/48ce0344eff94c8771c105165ea128ed.png)
, prove that we can find
![a_n](/media/m/1/f/f/1ff6f81c68b9c6fb726845c9ce762d7a.png)
so that
![b_n>c](/media/m/e/d/a/edaea950d43f94c341ff4073fd676d53.png)
for all sufficiently large
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
.
%V0
The real numbers $a_0,a_1,a_2,\ldots$ satisfy $1=a_0\le a_1\le a_2\le\ldots. b_1,b_2,b_3,\ldots$ are defined by $b_n=\sum_{k=1}^n{1-{a_{k-1}\over a_k}\over\sqrt a_k}$.
a.) Prove that $0\le b_n<2$.
b.) Given $c$ satisfying $0\le c<2$, prove that we can find $a_n$ so that $b_n>c$ for all sufficiently large $n$.