9231 P11 - Nov 2014 - Q1
6375
Given that
\(u_{k}=\frac{1}{\sqrt{2 k-1}}-\frac{1}{\sqrt{2 k+1}},\)
express \(\sum_{k=13}^{n} u_{k}\) in terms of \(n\).
Deduce the value of \(\sum_{k=13}^{\infty} u_{k}\).
Solution
Answer: \(\sum_{k=13}^{n} u_k = \frac{1}{5}-\frac{1}{\sqrt{2n+1}}\).
Hence, \(\sum_{k=13}^{\infty} u_k = \frac{1}{5}\).
Write out the sum:
\(\left(\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{27}}\right)+\left(\frac{1}{\sqrt{27}}-\frac{1}{\sqrt{29}}\right)+\cdots+\left(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}\right).\)
This is a telescoping series, so all the middle terms cancel, leaving
\(\sum_{k=13}^{n} u_k = \frac{1}{\sqrt{25}}-\frac{1}{\sqrt{2n+1}} = \frac{1}{5}-\frac{1}{\sqrt{2n+1}}\).
Now let \(n \to \infty\). Since \(\frac{1}{\sqrt{2n+1}} \to 0\),
\(\sum_{k=13}^{\infty} u_k = \frac{1}{5}\).
