Find lim
First I tried by taking \ln y_n=\ln \frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}=\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln(n),
which dose not seems to take me anywhere. Then I tried to use squeeze theorem, since \frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}\geq 1, 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)=\sqrt x \ln x you have
f'(x)=\frac{\ln x}{2\sqrt x}-\frac1{\sqrt x}.
Now by mean value theorem
f(n+1)-f(n) = f'(\theta_n)
for some \theta_n such that $n<\theta_n
If we notice that \lim\limits_{x\to\infty} f'(x) = 0 we get that
\lim\limits_{n\to\infty} (\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln n) = \lim\limits_{n\to\infty} f'(\theta_n) = 0.
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