Euler derived the following identity $$ 1+a_{1}+a_{1}a_{2}+a_{1}a_{2}a_{3}+\cdots= \cfrac{1}{ 1- \cfrac{a_{1}}{ 1+a_{1}- \cfrac{a_{2}}{ 1+a_{2}- \cfrac{a_{3}}{ 1+a_{3} - \ddots}}}}\;\;\;\;\;\;\;(1) $$ where the LHS can be expressed as $$ 1+a_1(1+a_2(1+a_3(1+a_4(1+a_5(\cdots)))))\;\;\;\;\;\;\;\;\;\;\;\;(2) $$ Now suppose that we replace those $1$'s with $b_{i}$'s we get $$ 1+a_{1}(b_{1}+a_{2}(b_{2}+a_3(b_{3}+a_4(b_{4}+a_5(\cdots)))))\;\;\;\;\;\;\;\;\;\;\;\;(3) $$ My question is: Is it possible to express $(3)$ as an infinite continued fraction?
Thanks.
The answer is easier than I initially thought, $(3)$ can be expressed in the form of $(2)$ as follows: $$ 1+a_{1}b_{1}\left(1+\frac{a_{2}b_{2}}{b_{1}}\left(1+\frac{a_{3}b_{3}}{b_{2}}\left(1+\frac{a_{4}b_{4}}{b_{4}}(1+\cdots)\right)\right)\right) $$ So the answer to the question is yes. $$ 1+a_{1}(b_{1}+a_{2}(b_{2}+a_3(b_{3}+a_4(\cdots)))= \cfrac{1}{ 1- \cfrac{a_{1}b_{1}}{ 1+a_{1}b_{1}- \cfrac{\frac{a_{2}b_{2}}{b_{1}}}{ 1+\frac{a_{2}b_{2}}{b_{1}}- \cfrac{\frac{a_{3}b_{3}}{b_{2}}}{ 1+\frac{a_{3}b_{3}}{b_{2}} - \ddots}}}} $$