# Finding $\lim_{n \to \infty}(n^{1\over n}+{1\over n})^{n\over \ln n}$

$$\lim_{n \to \infty}\left(n^{1\over n}+{1\over n}\right)^{n\over \ln n}$$

The limit is same as

$$e^{\displaystyle{\lim_{n \to \infty}}{n^{(1+{1\over n})}-n+1\over \ln n}}$$

But I am stuck here , I noticed that if I take $$n$$ common from the numerator then it is of the form $$0 \times \infty$$ , but I can't seem to solve further , using L'hopital's rule is too hectic or perhaps not meant to solve this one . I couldn't use any of the Maclaurin expansions either in this case .

Could someone please guide me in this ?

Thanks !

• Hint. You may write $$\left(n^{\frac{1}{n}}+\frac{1}{n}\right)^{\frac{n}{\log n}}=\exp\left( 1 + \frac{n}{\log n}\log\left(1+\frac{1}{n^{1+1/n}}\right)\right).$$ – Sangchul Lee Jun 30 '20 at 4:49
• Take $\log$ sides and note that $\lim n^{\frac{1}{n}}$ is same as $\lim n^{\frac{1}{n}} +\frac{1}{n}$ – Hrishabh Jun 30 '20 at 4:53
• @Hrishabh that was great , it got solved without even using the pen , thanks ! – ARROW Jun 30 '20 at 5:31

Here is simple approach for this limit. $$\lim_{n\to \infty} \left(\sqrt[n]{n} +\frac{1}{n} \right)^{\frac{n}{\ln n}} = \lim_{n\to\infty} \left[ n^{\frac{1}{\ln n}}\left(1+\frac{1}{n \sqrt[n]{n}}\right)^{\frac{n}{\ln n}}\right]=\lim_{n\to\infty} n^{\frac{1}{\ln n}} \exp\left(\lim_{n\to\infty} \frac{1}{\sqrt[n]{n} \ln n}\right) =e \times 1 = e$$ Since $$\lim_{n\to\infty} n^{\frac{1}{\ln n}} = \exp\left(\lim_{n\to \infty} \frac{\ln n}{\ln n}\right)=e$$

In the latter limit we use the fact that $$\displaystyle \sqrt[n]{n} =1$$ so as $$n\to \infty$$ and $$\ln n\to \infty$$ which tell us that $$\frac{1}{\sqrt[n]{n} \ln n}$$ is decreasing sequence for all $$n>1$$ and it's limit around large $$n$$ is $$\displaystyle \lim_{n\to\infty}\frac{1}{\ln n} =0$$ .

• I think that $\sqrt[n]{n}$ should be in the denominator in the last step .(right before $e \times 1$ ) – ARROW Jun 30 '20 at 5:37
• @ARROW, its typo. I have fixed it. Thank you for pointing out. :) – Naren Jun 30 '20 at 5:45

$$a_n=\left(n^{1\over n}+{1\over n}\right)^{n\over \log( n)}\implies \log(a_n)={n\over \log( n)}\log\left(n^{1\over n}+{1\over n}\right)$$ By Taylor $$\log\left(n^{1\over n}+{1\over n}\right)=\frac{\log (n)+1}{n}-\frac{2\log (n)+1}{2n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(a_n)=1+\frac 1 {\log(n)}-\frac{2\log (n)+1}{2n\log(n)}+O\left(\frac{1}{n^2}\right)$$ $$a_n=e^{\log(a_n)}=e \, \exp\left(\frac 1 {\log(n)}-\frac{2\log (n)+1}{2n\log(n)}+O\left(\frac{1}{n^2}\right) \right)$$ xhish shows the limit and how it is approached.

Moreover, this gives you a shortcut method for an estimate of $$a_n$$. Suppose $$n=10$$; the truncated expression would give $$3.716$$ while the exact value is $$3.788$$.

It would give a relative error lower than $$1$$% for $$n \geq 15$$ and lower than $$0.1$$% for $$n \geq 54$$.

Your first step is correct, as in taking the exponential and the logarithm.

However, the next quantity is not correct. Take $$L$$ to be the limit in question, then

$$L=\exp\left(\lim_{n\to\infty}\frac{n}{\log n}\log\left(n^{\frac1n}+\frac1n\right)\right).$$ I can't quite piece how did you get to your quantity but that's beside the point. Now rewrite $$\frac1n=n^{-1+\left(\frac1n-\frac1n\right)}=n^{\frac1n}\cdot n^{-\left(1+\frac1n\right)}.$$ Using the product rule for logarithms, the $$\log$$ right there becomes

$$\log\left(n^{\frac1n}+\frac1n\right)=\log\left(n^{\frac1n}+n^{\frac1n}\cdot n^{-\left(1+\frac1n\right)}\right)=\frac{\log n}n+\log\left(1+n^{-\left(1+\frac1n\right)}\right).$$ Back in the original expression we substitute back to get \begin{align*} L&=\exp\left(\lim_{n\to\infty}\frac{n}{\log n}\left(\frac{\log n}n+\log\left(1+n^{-\left(1+\frac1n\right)}\right)\right)\right)\\ &=\exp\left(1+\lim_{n\to\infty}\frac{n}{\log n}\cdot\log\left(1+n^{-\left(1+\frac1n\right)}\right)\right). \end{align*}

We now treat the last logarithm, so as an strategy, we know that $$\frac{n}{\log n}$$ goes to $$\infty$$ as $$n$$ grows since $$n\gg\log n$$. Therefore its inverse goes to zero, then we would like the other quantity to go to zero as well to apply L'Hôpital's rule. It is enough to show that $$n^{-\left(1+\frac1n\right)}$$ tends to zero as $$n$$ grows.

$$\lim_{n\to\infty}n^{-\left(1+\frac1n\right)}=\exp-\left(\lim_{n\to\infty}\left(1+\frac1n\right)\log(n)\right)=\exp-((1+0)\infty)=0.$$

However, I feel that trying to apply L'Hôpital's rule here would be too convoluted. If anyone is willing to expand on this procedure I would be grateful.

I'll be back once I see how to get around this.