Tuesday, 19 September 2017

Evaluate the limit mathoplimlimitsntoinftyfrac(n+1)log2(n+1)nlog2nlog2n



Evaluate:
limn(n+1)log2(n+1)nlog2nlog2n




Intuitively, I feel that for large n, log(n+1) log(n). So, the above limit should reduce to:



=limn{(n+1)n}log2nlog2n =1



However, can someone please suggest how can one mathematically show this.



Thanks!


Answer




If you want to be strict, write log(1+n)=log(n)+log(1+1n)

So, the numerator is A=(n+1)log2(n+1)nlog2(n)=(n+1)(log(n)+log(1+1n))2nlog2(n)
Expanding the square and grouping A=a2n+a2+2anlog(n)+2alog(n)+log2(n)
where a=log(1+1n)1n is small when n is large.



Is this what you are looking for ?



Edit



Alternatively, you could set n=1x and use Taylor series around x=0. Doing so, the entire expression will be (12log(x))+x(1log2(x)1log(x))+x23log(x)+O(x3)


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