This actually isn't a telescoping series, even though it appears to be one. You can rewrite it as
ln((1/2)(2/3)(3/4)...) = ln(1/n)
As n -> ∞, 1/n -> 0, and ln(0) is undefined. So the sum doesn't converge.
Even if you broke it up so it did telescope:
ln(1/2) + ln(2/3) + ... = ln(1) - ln(2) + ln(2) - ln(3) ... = ln(1) - ln(n) = -ln(n)
as n -> ∞, -ln(n) -> -∞ so the sum doesn't converge.
ln((1/2)(2/3)(3/4)...) = ln(1/n)
As n -> ∞, 1/n -> 0, and ln(0) is undefined. So the sum doesn't converge.
Even if you broke it up so it did telescope:
ln(1/2) + ln(2/3) + ... = ln(1) - ln(2) + ln(2) - ln(3) ... = ln(1) - ln(n) = -ln(n)
as n -> ∞, -ln(n) -> -∞ so the sum doesn't converge.
-
......... n ... .. .. .. .. .. .. .. . n
S(n) = ∑ ln( k/(k+1) ) = ln ( П ( k/(k+1) ) ) = ln(1/n+1)
......... k=1 .. .. .. .. .. .. .. . k=1
S(n) = ∑ ln( k/(k+1) ) = ln ( П ( k/(k+1) ) ) = ln(1/n+1)
......... k=1 .. .. .. .. .. .. .. . k=1