show that limn-->inff 1/(lnn)(1+1/2+.............+1/n^2)=2

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You may use Stolz-Cesaro's theorem such that:

`if lim_(n-gtoo) (x_(n+1)-x_n)/(y_(n+1) - y_n) = a =gt lim_(n-gtoo) (x_n)/(y_n) = a`

Hence, using this property yields:

`lim_(n-gtoo)(1+1/2+.....+1/n^2+1/((n+1)^2) - 1-1/2-....-1/n^2)/(ln(n+1)-ln(n))`

`lim_(n-gtoo)(1/((n+1)^2))/(ln ((n+1)/n)) = lim_(n-gtoo) 1/((n+1)^2*ln ((n+1)/n))`


`lim_(n-gtoo) 1 / (ln ((1+1/n)^((n+1)^2)))`

`lim_(n-gtoo) ln ((1+1/n)^((n+1)^2)) = ln lim_(n-gtoo) ((1+1/n)^((n+1)^2))`

`ln lim_(n-gtoo) ((1+1/n)^((n+1)^2)) = ln lim_(n-gtoo) ((1+1/n)^n)^((n+1)^2)/n) = ln e^oo = oo`

`lim_(n-gtoo) 1/(ln ((1+1/n)^((n+1)^2)) = 0`

Hence, evaluating the limit yields that `lim_(n-gtoo)(1+1/2+.....+1/n^2)/(ln n)!= 2`  but `lim_(n-gtoo)(1+1/2+.....+1/n^2)/(ln n) = 0.`

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