Consider writing out the terms of − ln a 1 + ∑ n = 1 k ln ( a n a n + 1 ) {\displaystyle -\ln a_{1}+\sum _{n=1}^{k}\ln({\frac {a_{n}}{a_{n+1}}})} for a general k. You may find it useful to recall ln ( x / y ) = ln x − ln y {\displaystyle \ln(x/y)=\ln x-\ln y} . You should find many terms cancel.
