Exercise 2.3.6

Question

Consider the sequence given by $b_{n}=n-\sqrt{n^{2}+2 n}$. Taking $(1 / n) \rightarrow 0$ as given, and using both the Algebraic Limit Theorem and the result in Exercise 2.3.1, show $\lim b_{n}$ exists and find the value of the limit.

Ulisse Mini · Accepted Answer

I'm going to find the value of the limit before proving it. We have
  $$
  n - \sqrt{n^2 + 2n} = n - \sqrt{(n + 1)^2 - 1}
  $$
  For large $n$, $\sqrt{(n + 1)^2 - 1} \approx n + 1$ so $\lim b_n = -1$.

Factoring out $n$ we get $n\left(1 - \sqrt{1 + 2/n}ight)$. Tempting as it is to apply the ALT here to say $(b_n) 	o 0$ it doesn't work since $n$ diverges.

How about if I get rid of the radical, then use the ALT to go back to what we had before?
  $$
  (n - \sqrt{n^2 + 2n})(n + \sqrt{n^2 + 2n}) = n^2 - (n^2 + 2n) = -2n
  $$
  Then we have
  $$
  b_n = n - \sqrt{n^2 + 2n} = \frac{-2n}{n + \sqrt{n^2 + 2n}} = \frac{-2}{1 + \sqrt{1 + 2/n}}
  $$
  Now we can finally use the algebraic limit theorem!
  $$
  \lim\left(\frac{-2}{1 + \sqrt{1 + 2/n}}ight) = \frac{-2}{1 + \sqrt{1 + \lim\left(2/night)}} = \frac{-2}{1 + \sqrt{1 + 0}} = -1
  $$

Stepping back the key to this technique is removing the radicals via a difference of squares, then dividing both sides by the growth rate $n$ and applying the ALT.

Exercise 2.3.6

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Exercise 2.3.6

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