Math
posted by Hannah .
Evaluate sum(n=2,n=infinity) ln(11/n^2) if it exists.

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11/n^2 = (n^2  1)/n^2 =(n+1)(n1)/n^2
>
Log(11/n^2) =
Log(n+1) + Log(n1)  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(11/n^2) has the same structure. We see that:
Log(11/n^2) = Delta^2 f(n)
with:
f(n) = log(n1)
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(s1) + g(s+1)  g(s) =
g(s+1)  g(r)
So, we have:
Sum from n = 2 to N of Log(11/n^2) =
Sum from n = 2 to N of
Delta(Delta f)(n) =
(Delta f)(N+1)  (Delta f)(2)
Now
f(n) = log(n1),
so
(Delta f)(n) = Log(n)  log(n1) =
Log[n/(n1)]
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)
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