YOU SAID:

{\displaystyle \prod _{n=1}^{\infty }\left[{1+{1 \over n(n+2)}}\right]^{\log _{2}(n)}}

INTO JAPANESE

{\displaystyle \prod _{n=1}^{\infty }\left[{1+{1 \over n(n+2)}}\right]^{\log _{2}(n)}}

BACK INTO ENGLISH

{\displaystyle \prod _{n=1}^{\infty }\left[{1+{1 \over n(n+2)}}\right]^{\log _{2}(n)}}

INTO JAPANESE

{\displaystyle \prod _{n=1}^{\infty }\left[{1+{1 \over n(n+2)}}\right]^{\log _{2}(n)}}

BACK INTO ENGLISH

{\displaystyle \prod _{n=1}^{\infty }\left[{1+{1 \over n(n+2)}}\right]^{\log _{2}(n)}}