2004 Paper 3 Q3

Year: 2004
Paper: 3
Question Number: 3

Course: UFM Pure
Section: Sequences and series, recurrence and convergence

Difficulty: 1700.0 Banger: 1516.0

Problem

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 \[ {3 \over 2} \displaystyle < \sum_{n=1}^\infty {1 \over n^2} <{7 \over 4}\;. \]

No solution available for this problem.

Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1516.0

Banger Comparisons: 1

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Problem source
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
\[
 {3 \over 2} \displaystyle 
< \sum_{n=1}^\infty {1 \over n^2} <{7 \over 4}\;.
\]