Thursday, 1 May 2014

sequences and series - Find the limit of frac(n+1)sqrtn+1nsqrtn.



Find limn(n+1)n+1nn



First I tried by taking lnyn=ln(n+1)n+1nn=n+1ln(n+1)nln(n),



which dose not seems to take me anywhere. Then I tried to use squeeze theorem, since (n+1)n+1nn1, but I need an upper bound now. It's for a while I am trying to come up with it but stuck. Can you help me please.


Answer




Notice that for f(x)=xlnx you have
f(x)=lnx2x1x.


Now by mean value theorem
f(n+1)f(n)=f(θn)

for some θn such that $n<\theta_n

If we notice that limxf(x)=0 we get that
limn(n+1ln(n+1)nlnn)=limnf(θn)=0.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...