Year: 2004
Paper: 3
Question Number: 3
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
No solution available for this problem.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Given that $\f''(x) > 0$ when $a \le x \le b\,$,
explain with the aid of a sketch why
\[
(b-a) \, \f \Big( {a+b \over 2} \Big)
< \int^b_a \f(x) \, \mathrm{d}x
< (b-a) \, \displaystyle \frac{\f(a) + \f(b)}{2} \;.
\]
By choosing suitable $a$, $b$ and $\f(x)\,$, show that
\[
{4 \over (2n-1)^2} < {1 \over n-1} - {1 \over n}
< {1 \over 2} \l {1 \over n^2} + {1 \over (n-1)^2}\r \,,
\]
where $n$ is an integer greater than 1.
Deduce that
\[
4 \l {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \r
< 1
< {1 \over 2} +
\left( {1 \over 2^2} +{1 \over 3^2} + {1 \over 4^2} + \cdots \right)\,.
\]
Show that
\[
{1 \over 2} \l {1 \over 3^2}
+ {1 \over 4^2} + {1 \over 5^2} + \frac 1 {6^2} + \cdots \r
<
{1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2}
+ \cdots
\]
and hence show that
\[
{3 \over 2} \displaystyle
< \sum_{n=1}^\infty {1 \over n^2} <{7 \over 4}\;.
\]