Wednesday, 16 December 2015

calculus - Determine limntoinftynleft(1+(n+1)lnfracnn+1right)



Determine limnn(1+(n+1)lnnn+1)


I noticed the indeterminate case 0 and I tried to get them all under the ln, but it got more complicated and I reached another indeterminate form. The same happened when I tried to use Stolz-Cesaro.




EDIT: is there an elementary solution, without l'Hospital or Taylor series?


Answer



Let nn+1=ex. As n,x0 through negative values.The expression is ex1+xex(1ex)2.

Applying l'Hopital's Rule to 1+xex(1ex)2 we get a limit of 12 as x0. And the far left term ex in the expression goes to 1 as x0.


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