Iv'e been struggling with this one for a bit too long:

$$ a_n = \lim_{n \to \infty} n^\frac{2}{3}\cdot ( \sqrt{n-1} + \sqrt{n+1} -2\sqrt{n} )$$

What Iv'e tried so far was using the fact that the inner expression is equivalent to that:

$$ a_n = \lim_{n \to \infty} n^\frac{2}{3}\cdot ( \sqrt{n-1}-\sqrt{n} + \sqrt{n+1} -\sqrt{n} ) $$

Then I tried multiplying each of the expression by their conjugate and got:

$$ a_n = \lim_{n \to \infty} n^\frac{2}{3}\cdot ( \frac{1}{\sqrt{n+1} +\sqrt{n}} - \frac{1}{\sqrt{n-1} +\sqrt{n}} ) $$

But now I'm in a dead end. Since I have this annyoing $n^\frac{2}{3}$ outside of the brackets, each of my attemps to finalize this, ends up with the undefined expression of $(\infty\cdot0)$

I've thought about using the squeeze theorem some how, but didn't manage to connect the dots right.


  • $\begingroup$ Yes, misspeled...sorry $\endgroup$
    – MorKadosh
    Jan 18, 2016 at 21:24

3 Answers 3


Keep on going... the difference between the fractions is


which, by similar reasoning as before (diff between two squares...), produces


Now, as $n \to \infty$, the denominator behaves as $(2 \sqrt{n})^3 = 8 n^{3/2}$. Thus, $\lim_{n \to \infty} (-1/4) n^{-3/2} n^{2/3} = \cdots$? (Is the OP sure (s)he didn't mean $n^{3/2}$ in the numerator?)

  • $\begingroup$ Thanks! So I suppose it should be 0? since the power of the nominator is smaller than the denomiator? $\endgroup$
    – MorKadosh
    Jan 18, 2016 at 21:43
  • $\begingroup$ @MorKadosh: the way you have the limit stated, yes, the limit is zero. I wonder though if you hadn't written something backward. $\endgroup$
    – Ron Gordon
    Jan 18, 2016 at 21:45
  • $\begingroup$ Nope, this is the exact thing...thanks! $\endgroup$
    – MorKadosh
    Jan 18, 2016 at 21:47
  • $\begingroup$ Solid Answer! +1 And closing in on the 100-K club at which time all will be revealed and you will receive total consciousness ... which is nice. ;-)) $\endgroup$
    – Mark Viola
    Jan 18, 2016 at 23:45
  • $\begingroup$ @Dr.MV: Heh. I'd be happy with a mug. I think they used to give them out when you hit, like, 20K, but that was when only smart people populated the site. $\endgroup$
    – Ron Gordon
    Jan 18, 2016 at 23:48

Consider the corresponding function: if it has a limit at $\infty$, it will be the same as for the sequence. For the limit of the function we can do a substitution $x=1/t$, so we have to compute \begin{align} \lim_{t\to0^+} \frac{1}{t^{2/3}}\frac{\sqrt{1-t}+\sqrt{1+t}-2}{\sqrt{t}} &= \lim_{t\to0^+}\frac{1-t/2-t^2/8+o(t^2)+1+t/2-t^2/8+o(t^2)-2}{t^{7/6}} \\ &=\lim_{t\to0^+}\frac{-t^2/4+o(t^2)}{t^{7/6}}=0 \end{align}

If the exponent in the first factor is $3/2$, rather than $2/3$, the denominator would be $t^2$ and the limit would be $-1/4$.


I thought I would present a way forward that is different from the approach used in the OP. We note that from the Mean Value Theorem, there exists a number $\xi\in (x,x+1)$ and a number $\eta \in (x-1,x)$ such that

$$f(x+1)-f(x)-f'(x)=\frac12 f''(\xi) \tag 1$$


$$f(x-1)-f(x)+f'(x)=\frac12 f''(\eta) \tag 2$$

Adding $(1)$ and $(2)$ yields

$$f(x+1)-2f(x)+f(x-1)=\frac12\left(f''(\xi)+f''(\eta)\right) \tag 3$$

Now, letting $f(x)=\sqrt{x}$ in $(3)$ reveals that


Therefore, we have

$$n^{2/3}\left(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\right)=-\frac18 n^{2/3}\left(\xi^{-3/2}+\eta^{-3/2}\right)$$

since $n-1<\eta<n<\xi<n+1$. Taking the limit as $n\to \infty$ we see that the limit of interest is $0$


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