Answer
(a) ${s_n}=\mathop \sum \limits_{k = 1}^n \ln \dfrac{k}{{k + 1}} = - \ln \left( {n + 1} \right)$
The series diverges.
(b) ${s_n}=\ln \dfrac{1}{2} - \ln \dfrac{{n + 1}}{{n + 2}}$
The series converges and its sum is $-\ln 2$.
Work Step by Step
(a) We have the series:
$\mathop \sum \limits_{k = 1}^\infty \ln \dfrac{k}{{k + 1}} = \ln \dfrac{1}{2} + \ln \dfrac{2}{3} + \ln \dfrac{3}{4} + \cdot\cdot\cdot + \ln \dfrac{k}{{k + 1}} + \cdot\cdot\cdot$
1. Find a closed form of ${s_n}$
The series can be rewritten as
$\mathop \sum \limits_{k = 1}^\infty \ln \dfrac{k}{{k + 1}} = \left( {\ln 1 - \ln 2} \right) + \left( {\ln 2 - \ln 3} \right) + \left( {\ln 3 - \ln 4} \right) + \cdot\cdot\cdot + \left( {\ln k - \ln \left( {k + 1} \right)} \right) + \cdot\cdot\cdot$
We then obtain the $n$th partial sums:
${s_n} = \mathop \sum \limits_{k = 1}^n \ln \dfrac{k}{{k + 1}} = 0 + \left( { - \ln 2 + \ln 2} \right) + \left( { - \ln 3 + \ln 3} \right) + \cdot\cdot\cdot + \left( { - \ln n + \ln n} \right) - \ln \left( {n + 1} \right)$
Thus,
$\mathop \sum \limits_{k = 1}^n \ln \dfrac{k}{{k + 1}} = - \ln \left( {n + 1} \right)$
The right-hand side of the equality is a closed form of ${s_n}$.
2. Evaluate the limit
$\mathop {\lim }\limits_{n \to \infty } {s_n} = - \mathop {\lim }\limits_{n \to \infty } \ln \left( {n + 1} \right) = - \infty $
The limit does not exist. So, the series diverges.
(b) We have the series:
$\mathop \sum \limits_{k = 1}^\infty \ln \left( {1 - \dfrac{1}{{{{\left( {k + 1} \right)}^2}}}} \right) = \ln \left( {1 - \dfrac{1}{4}} \right) + \ln \left( {1 - \dfrac{1}{9}} \right) + \ln \left( {1 - \dfrac{1}{{16}}} \right) + \cdot\cdot\cdot + \ln \left( {1 - \dfrac{1}{{{{\left( {k + 1} \right)}^2}}}} \right) + \cdot\cdot\cdot$
1. Find a closed form of ${s_n}$
Write
$1 - \dfrac{1}{{{{\left( {k + 1} \right)}^2}}} = \dfrac{{{k^2} + 2k}}{{{{\left( {k + 1} \right)}^2}}} = \dfrac{{k\left( {k + 2} \right)}}{{{{\left( {k + 1} \right)}^2}}} = \dfrac{k}{{\left( {k + 1} \right)}}\cdot\dfrac{{k + 2}}{{\left( {k + 1} \right)}}$
So, the $n$th partial sums of the series can be rewritten as
${s_n} = \mathop \sum \limits_{k = 1}^n \ln \left( {1 - \dfrac{1}{{{{\left( {k + 1} \right)}^2}}}} \right) = \mathop \sum \limits_{k = 1}^n \ln \left( {\dfrac{k}{{\left( {k + 1} \right)}}\cdot\dfrac{{k + 2}}{{\left( {k + 1} \right)}}} \right) = \mathop \sum \limits_{k = 1}^n \left( {\ln \dfrac{k}{{k + 1}} + \ln \dfrac{{k + 2}}{{k + 1}}} \right)$
${s_n} = \mathop \sum \limits_{k = 1}^n \left( {\ln \dfrac{k}{{k + 1}} - \ln \dfrac{{k + 1}}{{k + 2}}} \right)$
${s_n} = \left( {\ln \dfrac{1}{2} - \ln \dfrac{2}{3}} \right) + \left( {\ln \dfrac{2}{3} - \ln \dfrac{3}{4}} \right) + \left( {\ln \dfrac{3}{4} - \ln \dfrac{4}{5}} \right) + \cdot\cdot\cdot + \left( {\ln \dfrac{n}{{n + 1}} - \ln \dfrac{{n + 1}}{{n + 2}}} \right)$
${s_n} = \ln \dfrac{1}{2} + \left( { - \ln \dfrac{2}{3} + \ln \dfrac{2}{3}} \right) + \left( { - \ln \dfrac{3}{4} + \ln \dfrac{3}{4}} \right) + \left( { - \ln \dfrac{4}{5} + \ln \dfrac{4}{5}} \right) + \cdot\cdot\cdot + \left( { - \ln \dfrac{n}{{n + 1}} + \ln \dfrac{n}{{n + 1}}} \right) - \ln \dfrac{{n + 1}}{{n + 2}}$
${s_n} = \ln \dfrac{1}{2} - \ln \dfrac{{n + 1}}{{n + 2}}$
The right-hand side of the equality is a closed form of ${s_n}$.
2. Evaluate the limit
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\ln \dfrac{1}{2} - \ln \dfrac{{n + 1}}{{n + 2}}} \right) = \ln \dfrac{1}{2} - \ln \left( {\mathop {\lim }\limits_{n \to \infty } \dfrac{{1 + \dfrac{1}{n}}}{{1 + \dfrac{2}{n}}}} \right)$
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \ln \dfrac{1}{2} - \ln 1 = \ln \dfrac{1}{2}$
Thus, the series converges and its sum is $\ln \frac{1}{2}=-\ln 2$.
