Problem 3

Let

1=a_0\le a_1\le a_2\le\cdots

be a sequence of real numbers. Define

b_n=\sum_{k=1}^n\left(1-\frac{a_{k-1}}{a_k}\right)\frac1{\sqrt{a_k}}

. (a) Prove that

0\le b_n<2

for every

n

. (b) Given any

b

with

0\le b<2

, prove that there is such a sequence for which

b_n>b

for infinitely many

n

Step 1 of 5: Rewrite each summand

In plain words

The complicated fraction is width times height in disguise.

b_n=\\sum_{k=1}^n(a_k-a_{k-1})a_k^{-3/2}

Detailed analysis

Since $1-\frac{a_{k-1}}{a_k}=\frac{a_k-a_{k-1}}{a_k}$ , each summand equals $(a_k-a_{k-1})a_k^{-3/2}$ .