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))2−nlog2(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 (1−2log(x))+x(1log2(x)−1log(x))+x23log(x)+O(x3)
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