Question

the value of log x+ log(1+1/x)+log(1+1/1+x)+log(1+1/2+x)+.............+log(1+1/(n-1+x))

Answer 1

i assume that not 1+1 over the rest

Answer 2

it's only under 1 not whole

Answer 3

the addition of logs refers to multiplication of arguments
\[log(a)+lob(b)=log(ab)\]

Answer 4

the division of logs referes to subtraction ... so i believe there is a telesoping part to thos

Answer 5

log x
+log (1+x)/(x)
+log (x+2)/(1+x)
+log(3+x)/(2+x)
+.............
+log(n+x)/(n-1+x)


log (x)
+log (1+x) -log(x)
+log (x+2) -log(1+x)
+log(3+x) -log(2+x)
+.............
+log(n+x)/(n-1+x)

log(n+x) this sums out to

Answer 6

or you could simplify of the terms :
(1+1/x) = (x+1)/x
1+1/(1+x) = (x+2)/(x+1)
1+1/(2+x) = (x+3)/(x+2)
1+1/(3+x) = (x+4)/(x+3)
...
...
1+1/(n-1+x) = (x+n)/(x+n-1)
then use the property of log, it becomes