Proof of Limit: Solving for the Limit of a Sequence with Proof Techniques

[courtrigrad]

Hello all

I need help with the following problems:

Prove that $$\lim_{x\rightarrow \infty} (\sqrt{n+1} - \sqrt n)(\sqrt {n+ \frac {1}{2}}) = \frac {1}{2}$$

I know that $$\lim_{x\rightarrow \infty} (\sqrt{n+1} - \sqrt n) = 0$$. Then why wouldn't the limit be 0?

Also another question (I posted this in another thread, but it died )

$$a_n = \sqrt {n+1} - \sqrt n$$ find three numbers $$N_1 , N_2, N_3$$ such that

$$a_n = \sqrt {n+1} - \sqrt n < \frac {1}{10}$$ for every $$n > N_1$$
$$a_n = \sqrt {n+1} - \sqrt n < \frac {1}{100}$$ for every $$n > N_2$$
$$a_n = \sqrt {n+1} - \sqrt n < \frac {1}{1000}$$ for every $$n > N_3$$

 

[Curious3141]

Reexpress $$\sqrt{n + a}$$ as $$\sqrt{n}\sqrt{1 + \frac{a}{n}$$ and use the general binomial expansion to simplify the series. You only need the first order approximation.

That is,

$$\sqrt{1 + \frac{a}{n}} \approx \left(1 + (\frac{1}{2})(\frac{a}{n})\right)$$

For the first part the a-values are 1 and $\frac{1}{2}$

Simplify the algebra, and you'll find the limit.

For the second part use the simplified expression found in the first part for the expression. You should be able to find a simple inequality in each case and solve accordingly. Post if you need further hints.

 

[Muzza]

Then why wouldn't the limit be 0?

Because sqrt(n + 1/2) isn't bounded.

Surely 1/n tends to 0 as n tends to infinity. But what's the limit of n * 1/n?

 

[courtrigrad]

ok so for the first part, the simplified expression I got was:

$$\sqrt n (\sqrt{1 + \frac {1}{n}} - \sqrt n)\sqrt n ( \sqrt{1+ \frac {1}{2n})$$

So $$(\sqrt{n+1} - n)( \sqrt {n + \frac {1}{2})$$

and this equals $$\frac {1}{2}$$ ?

hmm.. it looks like i went in circles

 

[Curious3141]

ok so for the first part, the simplified expression I got was:

$$\sqrt n (\sqrt{1 + \frac {1}{n}} - \sqrt n)\sqrt n ( \sqrt{1+ \frac {1}{2n})$$

So $$(\sqrt{n+1} - n)( \sqrt {n + \frac {1}{2})$$

and this equals $$\frac {1}{2}$$ ?

hmm.. it looks like i went in circles

Have you tried using the binomial expansion and simplifying as I advised ?

 

[courtrigrad]

do you mean $$\sqrt n (\sqrt{1 + \frac {1}{n}} - \sqrt n)\sqrt n ( \sqrt{1+ \frac {1}{2n}))^n$$ ?

NOTE: This is in the beginning of the book. I haven't learned about series expansions yet

Thanks

 

[learningphysics]

ok so for the first part, the simplified expression I got was:

$$\sqrt n (\sqrt{1 + \frac {1}{n}} - \sqrt n)\sqrt n ( \sqrt{1+ \frac {1}{2n})$$

I think there's a mistake here. I think it should be:

$$\sqrt n (\sqrt{1 + \frac {1}{n}} - 1)\sqrt n ( \sqrt{1+ \frac {1}{2n})$$

I rewrite as:
$$\frac{(\sqrt{1 + \frac {1}{n}} - 1) ( \sqrt{1+ \frac {1}{2n}})}{\frac{1}{n}}$$

Substitue x=1/n. So x->0 for the limit.

Then you can use L'Hopital's rule (since numerator and denominator go to zero), if you've covered that.

 

[courtrigrad]

thanks for the help. but how would you use the binomail expansion? Thats what's confusing me

Thanks

 

[learningphysics]

thanks for the help. but how would you use the binomail expansion? Thats what's confusing me

Thanks

Have a look at my previous post. Do you know l'Hopital's rule? I personally find this easier, as I tend to make mistakes when substituting the binomial expansion.

But you can also use binomial expansion:

$$\sqrt {1+x} = {(1+x)}^{(1/2)}= 1 + (1/2)x + [(1/2)((1/2) - 1)/2!]x^2 + ...$$

Under the first square root x=1/n, under the second x=1/(2n).

 

[courtrigrad]

Thanks a bunch

 

[courtrigrad]

ok so now that i have the binomial expansion, how would i simplify it?
Would I use $$\sqrt{1 + \frac{a}{n}} \approx \left(1 + (\frac{1}{2})(\frac{a}{n})\right)$$ which when we substitute equals:

$$(1 + \frac{1}{2n})(1 + \frac{1}{4n}) = (1 + \frac{1}{4n} + \frac{1}{2n} + \frac{1}{8n^2})(\sqrt n)$$

Thanks

 

[HallsofIvy]

If you don't know the "binomial expansion", you shouldn't be trying to use it for this problem!

$$\lim_{x\rightarrow \infty} (\sqrt{n+1} - \sqrt n)(\sqrt {n+ \frac {1}{2}}) = \frac {1}{2}$$

There's nothing wrong with going ahead and multiplying those:
$$\sqrt{(n+1)(n+\frac{1}{2}}-\sqrt{n(n+\frac{1}{2}}$$

$$\sqrt{n^2+ \frac{3}{2}n+ \frac{1}{2}}-\sqrt{n^2+ \frac{1}{2}n}$$

Now do the usual "trick" of rationalizing the numerator: multiply both numerator and denominator (which is 1) by
$$\sqrt{n^2+ \frac{3}{2}n+ \frac{1}{2}}+\sqrt{n^2+ \frac{1}{2}n}$$
That gives
$$\frac{n+ \frac{1}{2}}{\sqrt{n^2+ \frac{3}{2}n+ \frac{1}{2}}+\sqrt{n^2+ \frac{1}{2}n}}$$

Finally divide both numerator and denominator by n (which becomes n² inside the square roots) and take the limit as n goes to infinity.

For the second question, just go ahead and solve the equations!

Find N₁ such that $$a_n = \sqrt {n+1} - \sqrt n < \frac {1}{10}$$
for all n> N₁.

Well, to pass from "<1/10" to ">1/10" it has to be "= 1/10". When is
$$a_n = \sqrt {n+1} - \sqrt n = \frac {1}{10} ?$$

Write this as $\sqrt{n+1}= \sqrt{n}+ \frac{1}{10}$ and square both sides:

$n+ 1= n+ \frac{2\sqrt{n}}{10}+ \frac{1}{100}$.
Rewrite as $\frac{2\sqrt{n}}{10}= 1- \frac{1}{100}= \frac{99}{100}$ and square again: $\frac{4n}{100}= \frac{9801}{10000}$ so
$n= \frac{9801}{400}= 24.5$ (approx.). Since n has to be an integer, n must be larger than 24: N₁= 24.

Do the same for the others.

 

[courtrigrad]

thank you for your very helpful response HallsofIvy. Also I want to thank everyone that took the time to help me with this problem.

 

[courtrigrad]

just had a quick question"

for $$\frac{n+ \frac{1}{2}}{\sqrt{n^2+ \frac{3}{2}n+ \frac{1}{2}}+\sqrt{n^2+ \frac{1}{2}n}}$$ how do we get the denominator after we divide by n?

Thanks
EDIT: Nevermind I got it. You divide by n^2 because of the square root sign