Help guys i will medal. The question is attached in the comments section.

Question

anonymous · Answer

Here

amistre64 · Answer

$$[\frac12ln^2(x)]'=\frac1xln(x)$$

amistre64 · Answer

i get to 0*inf/2

anonymous · Answer

Proceed i mainly need the limit not the integration

anonymous · Answer

Yeeee same 0*infinity thenn!!!

anonymous · Answer

@amistre64

amistre64 · Answer

[ln^2(n+1)- ln^2(n)]/2

[(ln((n+1)/n))(ln(n^2+n))]/2

   ln(1) ln(inf)/2

amistre64 · Answer

the wolf agrees that it is 0 overall

anonymous · Answer

100% but then what comes? Thats were im stuck

anonymous · Answer

Yepp but howw

amistre64 · Answer

well

ln^2(n+1)-ln^2(n)  as n to infinity ... by observation we could determine that +1 is immaterial for large value of n ... ln^2-ln^2 = 0

amistre64 · Answer

if your wanting a rigorous approach, i have none that come to mind.  ganesh might

anonymous · Answer

Nahhh thats not clear

anonymous · Answer

Mmm thanks btw ill keep on trying :)

amistre64 · Answer

by a difference of squares i got to:

[(ln((n+1)/n))(ln(n^2+n))]/2

but yeah, i have no good ideas after that and just noticed that we approach ln(1) as a factor ...