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
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Log(1-1/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)
with:
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)
Now
f(n) = log(n-1),
so
(Delta f)(n) = Log(n) - log(n-1) =
Log[n/(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)
To evaluate the given sum, we will use the concept of infinite series and try to find the sum of the series.
The given series is
sum(n=2,n=infinity) ln(1-1/n^2)
To start, let's rewrite the term inside the natural logarithm using a logarithmic identity. Writing it as ln(1 - 1/n^2), we can apply the identity:
ln(1 - x) = - sum(n=1, n=infinity) (x^n) / n
where x is a real number with |x| < 1.
Applying this identity to our series, we get:
ln(1 - 1/n^2) = - sum(n=1, n=infinity) (1/n^2)^n / n
Now, let's simplify further by expanding the series. We can rewrite (1/n^2)^n as 1/n^(2n):
ln(1 - 1/n^2) = - sum(n=1, n=infinity) (1/n^(2n)) / n
Now, let's divide the numerator and the denominator by n^(2n). This will give us:
ln(1 - 1/n^2) = - sum(n=1, n=infinity) (1/n^(2n + 1))
Now, we observe that the denominator of the series contains both n and n^2 terms. Since n^2 grows faster than n, as n approaches infinity, the terms in the series will approach zero.
So, as n approaches infinity, the terms in the series (1/n^(2n + 1)) will tend to zero. Therefore, the sum of the series approaches zero.
In conclusion, the given series sum(n=2,n=infinity) ln(1-1/n^2) does exist and its value is zero.