# show that k=2 sigma k tends to infinity ln((1)-(1/k^2))=ln2

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### 1 Answer

You should bring the terms of difference `1 - 1/(k^2)` to a common denominator such that:

`Sigma_(k=2)^oo ln(1 - 1/(k^2)) = Sigma_(k=2)^ooln((k^2 - 1)/(k^2))`

`Sigma_(k=2)^oo (ln(k^2 - 1) - ln k^2) = Sigma_(k=2)^oo ln(k^2 - 1) - Sigma_(k=2)^oo ln k^2`

`Sigma_(k=2)^oo ln(k^2 - 1) - Sigma_(k=2)^oo ln k^2= ln(2^2 - 1) + ln(3^2 - 1) + ln(4^2 - 1) + ... + ln(oo) - ln 2^2 - ln 3^2 - .... ln oo`

`Sigma_(k=2)^oo ln(k^2 - 1) - Sigma_(k=2)^oo ln k^2 = ln 3 + ln 8 + ln 15 + .... + ln oo - ln 4 - ln 9 - ln 16 - ... - ln oo`

`Sigma_(k=2)^oo ln(k^2 - 1) - Sigma_(k=2)^oo ln k^2 = ln(3/4) + ln (8/9) + ln(15/16) + ....+ ln(k/(k+1)) + ln oo`

Evaluating the summation yields:

`Sigma_(k=2)^oo ln(1 - 1/(k^2)) = Sigma_(k=2)^oo(k/(k+1)) `

`Sigma_(k=2)^oo (k/(k+1)) = lim_(n-gtoo) Sigma_(k=2)^n ln(k+5)/(k+6)`

`Sigma_(k=2)^(n+5) ln(k+5)/(k+6) = ln 2 - ln 3 + ln 3 - ln 4 + ln 4 - ln 5 + .... + ln (n-1) - ln n + ln n - ln (n+1) + ... ln (n+5)/(n+6)`

`Sigma_(k=2)^n ln(k+5)/(k+6) = ln 2 - ln (n+1)`

`lim_(n-gtoo)Sigma_(k=2)^(n+5) ln(k+5)/(k+6) = lim_(n-gtoo) ln 2 - lim_(n-gtoo) ln (n+5)/(n+6)`

`lim_(n-gtoo)Sigma_(k=2)^(n+5)ln(k+5)/(k+6) = ln 2 - ln lim_(n-gtoo) (n(1+5/n))/(n(1+6/n))`

`lim_(n-gtoo) Sigma_(k=2)^(n+5)ln(k+5)/(k+6) = ln 2 - ln 1`

`lim_(n-gtoo) Sigma_(k=2)^(n+5)ln(k+5)/(k+6) = ln 2 - 0`

`lim_(n-gtoo) Sigma_(k=2)^(n+5)ln(k+5)/(k+6) = ln 2`

**Hence, evaluating the given summation yields `Sigma_(k=2)^oo ln(1 - 1/(k^2)) = ln 2.` **