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...

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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.**