(a) Consider the sum \(\sum_{r=1}^{n} \left( \ln r - 2 \ln (r+1) + \ln (r+2) \right)\).

This expands to:

• \(\ln 1 - 2 \ln 2 + \ln 3\)
• \(\ln 2 - 2 \ln 3 + \ln 4\)
• \(\ln 3 - 2 \ln 4 + \ln 5\)
• \(\ldots\)
• \(\ln (n-1) - 2 \ln n + \ln (n+1)\)
• \(\ln n - 2 \ln (n+1) + \ln (n+2)\)

Most terms cancel, leaving:

\([\ln 1] - \ln 2 - \ln (n+1) + \ln (n+2) = \ln \frac{n+2}{2(n+1)}\).

(b) The sum to infinity is \(S = -\ln 2\).

We need \(S_n - S = \ln \left( \frac{n+2}{n+1} \right) < 0.01\).

This leads to \(n+2 < e^{0.01}(n+1)\).

Solving gives the least value of \(n\) as 99.