What is the Limit of the Sequence b_n = n - sqrt(n^2 + 2n)?

[QuantumP7]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

Does the (1/n) \rightarrow 0 imply that I should put b_n 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.

 

[Samy_A]

\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.

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

 

[QuantumP7]

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

 

[Ray Vickson]

Homework Statement

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}) \nrightarrow 0), show \lim b_{n} exists and find the value of the limit.

Homework Equations

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

The Attempt at a Solution

Does the (1/n) \rightarrow 0 imply that I should put b_n 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.

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.

 

[QuantumP7]

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.

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!

 

[MidgetDwarf]

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