# Limit of a sequence problem

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1. Jan 4, 2016

### QuantumP7

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

2. Relevant equations
$b_{n} = n - \sqrt{n^{2} + 2n}$ and $(1/n) \rightarrow 0$

3. The attempt at a solution
Does the $(1/n) \rightarrow 0$ imply that I should put bn in the form 1/n? Going in that direction, I'm stuck at $$\frac{-2n}{n + \sqrt{n^{2} + 2n}}$$ Am I going in the right direction? And if so, any hints on how to further manipulate what I have? I'm self-studying, and do not have a professor or mentor or anything to give me a bit of direction.

2. Jan 4, 2016

### Samy_A

That certainly seems a right direction. Maybe divide both numerator and denominator by $n$, and take the $\frac{1}{n}$ into the square root.

3. Jan 4, 2016

### QuantumP7

YES! That is the hint I needed! Thank you SO much!!!

4. Jan 4, 2016

### Ray Vickson

You can write $\sqrt{n^2+2n} = \sqrt{n^2} \, \sqrt{1 + (2/n)} = n \sqrt{1 + (2/n)}$. Since you know that $\lim_{n \to \infty} \sqrt{1+(2/n)} = 1$, it follows that for any (small) number $r > 0$ we have $1 \leq \sqrt{1+(2/n)} < 1 + r$ for all $n$ larger than some number $N = N_r$. Therefore, for all $n> N$ we have that the denominator of your ratio above lies between $n + n = 2n$ and $n + n(1+r) = n(2+r)$. Therefore the ratio itself lies between $-2/2 = -1$ and $-2/2(1+r/2) = -1/(1 + (r/2))$. Therefore, the limit, $L$ is a number that is $\geq -1$ and $\leq -1/(1+(r/2))$, no matter how small the positive number $r$ that you choose. For example, if $r = 0.00002$ we have $-1 \leq L <-1/1.00001$. If $r = .00000000002$ then we have $-1 \leq L < 1/1.00000000001$, etc. Basically, $L$ must be -1; there is no other possible choice.

Last edited: Jan 4, 2016
5. Jan 5, 2016

### QuantumP7

Thank you SO much! I hadn't even thought to use $\sqrt{n^{2} + 2n} = \sqrt{n^{2}} \sqrt{1 + \frac{2}{n}}$!
Using my fraction, I somehow got that the limit of $b_{n}$ was -2. I see where my mistake was. When I used $$\frac{\frac{-2n}{n}}{\frac{n + \sqrt{n^{2} + 2n}}{n}} = \frac{-2}{1 + \frac{\sqrt{n^2 + 2n}}{n}}$$ and then setting the limit as $n \rightarrow \infty$, I made the mistake of assuming that $lim_{n \to \infty} \frac{\sqrt{n^{2} + 2n}}{n} = 0$ when that is not true.
Doing it your way, I have $lim_{n \to \infty} b_{n}$ = -1, as do you.

Thank you so much!!
The things you learn when you reach out for help!

6. Jan 9, 2016

### MidgetDwarf

You can also do it by induction, much harder and not needed here.