posted by Hannah on .
Evaluate sum(n=2,n=infinity) ln(1-1/n^2) if it exists.
1-1/n^2 = (n^2 - 1)/n^2 =(n+1)(n-1)/n^2
Log(n+1) + Log(n-1) - 2 Log(n).
If we define the difference operator Delta as:
(Delta f)(n) = f(n+1) - f(n)
Then applying Delta twice gives:
(Delta^2 f)(n) =
f(n+2) - 2f(n+1) + f(n)
The formula for Log(1-1/n^2) has the same structure. We see that:
Log(1-1/n^2) = Delta^2 f(n)
f(n) = log(n-1)
In general, we have:
Sum from n = r to s of (Delta g)(n) =
g(r+1) - g(r) + g(r+2) - g(r+1) + ...
+ g(s) - g(s-1) + g(s+1) - g(s) =
g(s+1) - g(r)
So, we have:
Sum from n = 2 to N of Log(1-1/n^2) =
Sum from n = 2 to N of
Delta(Delta f)(n) =
(Delta f)(N+1) - (Delta f)(2)
f(n) = log(n-1),
(Delta f)(n) = Log(n) - log(n-1) =
and we see that:
(Delta f)(N+1) - (Delta f)(2) =
Log[(N+1)/N] - Log(2)
The sum to infinity is the limit for N to infinity. We can write:
Log[(N+1)/N] = Log(1+1/N) =
1/N + O(1/N^2)
So, this terms goes to zero in the limit N to infinity.
The summation is thus equal to -Log(2)