Answers
ln ( n + 1 n ) − ln 2. \ln \left( \frac{n+1}{n} \right) - \ln 2. ln ( n n + 1 ) − ln 2.
ln 189 200 . \ln \frac{189}{200}. ln 200 189 . Full solutions
(a)
∑ r = 2 n ln ( ( r − 1 ) ( r + 1 ) r 2 ) = ∑ r = 2 n ( ln ( r − 1 ) − 2 ln r + ln ( r + 1 ) ) = ln 1 − 2 ln 2 + ln 3 + ln 2 − 2 ln 3 + ln 4 + ln 3 − 2 ln 4 + ln 5 + ⋯ + ln ( n − 3 ) − 2 ln ( n − 2 ) + ln ( n − 1 ) + ln ( n − 2 ) − 2 ln ( n − 1 ) + ln n + ln ( n − 1 ) − 2 ln n + ln ( n + 1 ) = − ln 2 − ln n + ln ( n + 1 ) = ln ( n + 1 n ) − ln 2 ■ \begin{align*}
& \sum_{r=2}^n \ln \left( \frac{(r-1)(r+1)}{r^2} \right) \\
& = \sum_{r=2}^n \left( \ln (r-1) - 2 \ln r + \ln (r+1) \right) \\
& = \def\arraystretch{1.5}
\begin{array}{lclclc}
& \ln 1 &-& 2 \ln 2 &+& \cancel{\ln 3} \\
+ & \ln 2 &-& \cancel{2 \ln 3} &+& \cancel{\ln 4} \\
+ & \cancel{\ln 3} &-& \cancel{2 \ln 4} &+& \cancel{\ln 5} \\
+ & \cdots & & & & \\
+ & \cancel{\ln (n-3)} &-& 2 \cancel{\ln (n-2)} &+& \cancel{\ln (n-1)} \\
+ & \cancel{\ln (n-2)} &-& 2 \cancel{\ln (n-1)} &+& \ln n \\
+ & \cancel{\ln (n-1)} &-& 2 \ln n &+& \ln (n+1) \\
\end{array} \\
& = - \ln 2 - \ln n + \ln (n+1) \\
& = \ln \left( \frac{n+1}{n} \right) - \ln 2 \; \blacksquare
\end{align*} r = 2 ∑ n ln ( r 2 ( r − 1 ) ( r + 1 ) ) = r = 2 ∑ n ( ln ( r − 1 ) − 2 ln r + ln ( r + 1 ) ) = + + + + + + ln 1 ln 2 ln 3 ⋯ ln ( n − 3 ) ln ( n − 2 ) ln ( n − 1 ) − − − − − − 2 ln 2 2 ln 3 2 ln 4 2 ln ( n − 2 ) 2 ln ( n − 1 ) 2 ln n + + + + + + ln 3 ln 4 ln 5 ln ( n − 1 ) ln n ln ( n + 1 ) = − ln 2 − ln n + ln ( n + 1 ) = ln ( n n + 1 ) − ln 2 ■
(b)
ln ( n + 1 n ) − ln 2 = ln ( 1 + 1 n ) − ln 2 \begin{align*}
& \ln \left( \frac{n+1}{n} \right) - \ln 2 \\
& = \ln \left( 1 + \frac{1}{n} \right) - \ln 2
\end{align*} ln ( n n + 1 ) − ln 2 = ln ( 1 + n 1 ) − ln 2 As n → ∞ , n \to \infty, n → ∞ , 1 n → 0 \frac{1}{n} \to 0 n 1 → 0 ln ( 1 + 1 n ) − ln 2 → − ln 2 \ln \left( 1 + \frac{1}{n} \right) - \ln 2 \to - \ln 2 ln ( 1 + n 1 ) − ln 2 → − ln 2
Hence the infinite series converges
and the sum to infinity is − ln 2 ■ - \ln 2 \; \blacksquare − ln 2 ■
(c)
∑ r = 10 20 ln ( ( r − 1 ) ( r + 1 ) r 2 ) = ∑ r = 2 20 ln ( ( r − 1 ) ( r + 1 ) r 2 ) − ∑ r = 2 9 ln ( ( r − 1 ) ( r + 1 ) r 2 ) = ln ( 20 + 1 20 ) − ln 2 − ( ln ( 9 + 1 20 ) − ln 2 ) = ln ( 21 20 ÷ 10 9 ) = ln 189 200 ■ \begin{align*}
& \sum_{r=10}^{20} \ln \left( \frac{(r-1)(r+1)}{r^2} \right) \\
& = \sum_{r=2}^{20} \ln \left( \frac{(r-1)(r+1)}{r^2} \right) - \sum_{r=2}^{9} \ln \left( \frac{(r-1)(r+1)}{r^2} \right) \\
& = \ln \left( \frac{20+1}{20}\right) - \ln 2 - \left( \ln \left( \frac{9+1}{20} \right) - \ln 2 \right) \\
& = \ln \left( \frac{21}{20} \div \frac{10}{9} \right) \\
& = \ln \frac{189}{200} \; \blacksquare
\end{align*} r = 10 ∑ 20 ln ( r 2 ( r − 1 ) ( r + 1 ) ) = r = 2 ∑ 20 ln ( r 2 ( r − 1 ) ( r + 1 ) ) − r = 2 ∑ 9 ln ( r 2 ( r − 1 ) ( r + 1 ) ) = ln ( 20 20 + 1 ) − ln 2 − ( ln ( 20 9 + 1 ) − ln 2 ) = ln ( 20 21 ÷ 9 10 ) = ln 200 189 ■
Question Commentary
Method of differences is the way to go for this question where we apply our
logarithm rules to get the differences. Subsequently we have the common follow-up
concepts of convergence, sum to infinity, and modifying the limits of a sum.