How can it be proved that: Sigma (k=2 to infinity) [ln(k-1) + ln(k+1) - 2*ln k] = -ln 2

We have to prove that Sigma (k = 2 to inf) [ ln(k-1) +
ln(k+1) - 2 *ln k]= - ln (2)

Sigma (k = 2 to inf) [ ln(k-1)
+ ln(k+1) - 2*ln k]

Now for k = 2 to inf., we
have

ln(k-1) + ln(k+1) - 2*ln k + ln(k-1) + ln(k+1) - 2*ln
k+ ln(k-1) + ln(k+1) - 2*ln k +...

=> ln 1 +
ln 3 - 2*ln 2 + ln 2 + ln 4 - 2*ln 3 +
ln 3 + ln 5 - 2*ln 4 + ln 4 + ln 6 - 2*ln 5 + ln 5 + ln 6 - 2*ln
6...inf

As the series above goes on till infinity, it can
be seen that all terms can be canceled except ln 1 and ln 2. For example I have
italicised the terms with ln 3. We can find similar sets for all the greater
numbers.

=> ln 1 - ln
2

as ln 1 = 0

=> -ln
2

This means we are only left with -ln 2 and this is what
we have to prove.

The required identity has
been proved.