What is the sum of the series ##1+ln2+(((ln2)^2)/(2!))+…+(((ln2)^2)/(n!))+…##?

I think that should be ##1 + ln2+(ln2)^2/(2!)+ . . .+(ln2)^n/(n!)+ . . .## if that is the case Simply apply the Maclaurin Series

##e^x = 1 + x + x^2/(2!)+x^3/(3!) + . . .+x^n/(n!) + . . .##

then putting x = ln2 the sum is:

##e^ln2 = 2##

and if it is the real question apply algebraic manipulation

##1 + ln2+(ln2)^2/(2!)+ . . .+(ln2)^2/(n!)+ . . . = 1 +ln2 +(ln2)^2[1/(2!)+1/(3!) + . . .+1/(n!) + . . .]##

putting x = 1

##e^1 = 1+ 1+1/(2!) + 1/(3!)+ . . .+1/(n!) + . . . =e## then the expression:

##1/(2!)+1/(3!) + . . .+1/(n!) + . . .## can be written as

or equal to ##e – 2## so the sum of

##1 + ln2+(ln2)^2/(2!)+ . . .+(ln2)^2/(n!)+ . . . = 1 +ln2 +(ln2)^2[1/(2!)+1/(3!) + . . .+1/(n!) + . . .] = 1+ln2+(ln2)^2(e-2)##

NOte: i think you had a typographical error.. that should be n not 2.