# calculate $\lim_{n\rightarrow\infty} n\left ( \ln(n^2 +1) -2\ln(n)\sqrt[n]{\ln(n)}\right )$

As above, I have to calculate $$\lim_{n\rightarrow\infty} n\left ( \ln(n^2 +1) -2\ln(n) \sqrt[n]{\ln(n)} \right )$$I tried to multiply it by $$\frac{\left ( \ln(n^2 +1) +2\ln(n)\sqrt[n]{\ln(n)}\right )}{\left ( \ln(n^2 +1) +2\ln(n)\sqrt[n]{\ln(n)}\right )}$$ so that it is $$\lim_{n\rightarrow\infty} n \frac{\left ( \ln^2(n^2 +1) -2\ln^2 (n) \sqrt[\frac{1}{2}n]{\ln(n)} \right )}{\left ( \ln(n^2 +1) +2\ln(n)\sqrt[n]{\ln(n)}\right )},$$but it is not so easy to play with (first glance says that it goes to infinity), nevertheless, can somebody take a closer look at this and suggest something, please? I would be very grateful.

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## 2 Answers

The following expansions yield readily an equivalent, hence the limit:

• $\ln(n^2+1)=2\log(n)+\log\left(1+\frac1{n^2}\right)=2\log(n)+o\left(\frac1n\right)$.
• $\sqrt[n]{\log n}=\exp\left(\frac1n\log\log n\right)=1+\frac1n\log(\log n)+o\left(\frac1n\right)$.
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$\ln(n^2+1)=2 \ln(n) +\ln (1+\frac{1}{n^2})$, so you got $2n\ln(n)(1-(\ln n)^{\frac{1}{n}}) +n \ln (1+\frac{1}{n^2})$.

The second summand goes to $0$ by Taylor expansion of $\ln(1+x)=x+O(x^2)$.

The first summand goes to $-\infty$ as $n \ln n (\ln n)^{\frac{1}{n}}-1> B$ is the same as $(\ln n)^{\frac{1}{n}}-1 > \frac{B}{n \ln n}$ or $(\ln n)^{\frac{1}{n}} > 1+ \frac{B}{n \ln n}$ or $\ln n > (1+ \frac{B}{n \ln n})^n$.

But for large $n$ we get

$\ln n > 2 e^{B} > (1+ \frac{B}{n})^n > (1+ \frac{B}{n \ln n})^n$, as wanted.

So pending any mistakes the whole thing goes to $-\infty$.

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