i cannot figure out a way to find this limit. $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$$

Its undeterminate form $0\cdot\infty$ so i tried using $$\displaystyle\lim_{n\to \infty}\frac{a^{X_n}-1}{X_n}=\ln{a}$$ this leads to $0\cdot\infty$ again. I then tried to transform it in $\frac{0}{0}$ $$\displaystyle\lim_{n\to \infty} \frac{\sqrt[n]{2}-\sqrt[n+1]{2}}{\frac{1}{n^2}}$$ here i dont know what i could do to find the limit.

  • $\begingroup$ Are you allowed to use Hopital's rule? I guess, you are not. $\endgroup$
    – Mikasa
    Jan 27 '13 at 10:45
  • $\begingroup$ Hint: $e^x=1+x+x^2/2+O(x^3)$ as $x\to0$. $\endgroup$
    – Hanul Jeon
    Jan 27 '13 at 10:45
  • $\begingroup$ I am not allowed to use Hopital's rule and i do not know big O notation. $\endgroup$
    – phi
    Jan 27 '13 at 10:46


$$2^{\frac{1}{n}} = e^{\frac{\log{2}}{n}}$$

so that

$$\begin{align} \lim_{n \rightarrow \infty} n^2 \left (2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right ) &= \lim_{n \rightarrow \infty} n^2 \left (e^{\frac{\log{2}}{n}} - e^{\frac{\log{2}}{n+1}} \right ) \\ &= \lim_{n \rightarrow \infty} n^2 \log{2} \left ( \frac{1}{n} - \frac{1}{n+1}\right ) \\ &= \lim_{n \rightarrow \infty} \frac{n \log{2}}{n+1} \\ &= \log{2} \\ \end{align} $$

The second step relies on the fact that

$$e^x = 1+ x + O(x^2)$$

as $x \rightarrow 0$.

  • $\begingroup$ You actually have to be a little more careful here: while it's true that the $O(x^2)$ terms cancel when you expand out $e^x$ for $x=\frac{\log 2}{n}$ and $x=\frac{\log 2}{n+1}$, that's not necessarily self-evident on the face of it, and of course because we're multiplying by $n^2$, if those second-order terms didn't cancel then they'd contribute a finite amount to the solution. Better to expand out to order 3. $\endgroup$ Jan 28 '13 at 3:24
  • $\begingroup$ @StevenStadnicki: You make a good point, although as you know the $k$th-order term provides a contribution of $O(n^{1-k})$ to the limit piece, which of course goes to zero in this limit when $k>1$. I could have been more explicit about that. $\endgroup$
    – Ron Gordon
    Jan 28 '13 at 10:04

By the mean value theorem $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})=\displaystyle\lim_{n\to \infty} \left(n^2 \times\frac{2^{1/c_{n}}\ln 2}{c_n^2}\right)=\ln2$$ where $n<c_n<n+1$


  • 1
    $\begingroup$ this is an outstanding solution $\endgroup$
    – user29743
    Jan 28 '13 at 2:33
  • $\begingroup$ @countinghaus To be fair, L'Hôpital is just a clever application of the intermediate value theorem too. $\endgroup$
    – Pedro Tamaroff
    Feb 27 '13 at 2:32

We have:

$ \begin{align*} \lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2}) & = \lim_{n\to \infty} n^2\left( 2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right) = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n}-\frac{1}{n+1}} - 1 \right) \\ & = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right) \\ & = \lim_{n\to \infty} \cfrac{n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right)}{\cfrac{1}{n^2+n}(n^2+n)} \\ & = \lim_{n\to \infty} \dfrac{n^2}{n^2+n} \cdot \lim_{n\to \infty} 2^{\frac{1}{n+1}} \cdot \lim_{n\to \infty} \cfrac{2^{\frac{1}{n^2+n}} - 1}{\cfrac{1}{n^2+n}} \\ & = 1 \cdot 1 \cdot \ln 2 \\ & = \boxed{\ln 2}. \end{align*} $


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