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 first few terms:
\(u_{13}=\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{27}},\quad u_{14}=\frac{1}{\sqrt{27}}-\frac{1}{\sqrt{29}},\quad u_{15}=\frac{1}{\sqrt{29}}-\frac{1}{\sqrt{31}}.\)
In general, the final term is
\(u_n=\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}.\)
So
\(\sum_{k=13}^{n} u_k=\left(\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{27}}\right)+\left(\frac{1}{\sqrt{27}}-\frac{1}{\sqrt{29}}\right)+\cdots+\left(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}\right).\)
This is a telescoping sum, so all the middle terms cancel:
\(\sum_{k=13}^{n} u_k=\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{2n+1}}=\frac{1}{5}-\frac{1}{\sqrt{2n+1}}.\)
Now let \(n\to\infty\). Since \(\frac{1}{\sqrt{2n+1}}\to 0\),
\(\sum_{k=13}^{\infty}u_k=\frac15.\)
