Limit of a fraction as n-> infinity in the numerator and denomominator

[Sparky_]

Homework Statement

Find limit an n-> infinity (n*sqrt(n)) / (n+1)sqrt(n+1))

Relevant Equations

L'Hopital's rule f'(n) / g'(n)

Hello,This is actually a piece of a little bigger problem (convergence of a series) - you can see the ratio test ak+1 / ak
That's why the (n) and (n+1) terms

I have lim n->∞ of (n√n) / (n+1)√(n+1) ∞/∞

I have tried L'Hopitals rule (requiring multiple times) and I am not seeing an end.

derivative of n^3/2/ (n(n+1)^1/2 + (n+1)^1/2)

lim 3/2 n^1/2 / (1)( n+1)^1/2 + (n)1/2(n+1)^(-1/2) + (1/2 (n+1)^(-1/2) )

form is still ∞/∞

Looking ahead ... it looks as if continuing to apply L'Hopitals rule because of the fractional exponents I will be stuck in a repetitive always getting ∞/∞

The answer (just the answer) was provided to this problem, the series does converge (within a bound) so I know this limit is not ∞/∞

Am I on the right track with L'Hopitals Rule or am I overlooking some simplification to make things easier?

Thanks
Sparky_

 

[PeroK]

Homework Statement:: Find limit an n-> infinity (n*sqrt(n)) / (n+1)sqrt(n+1))
Relevant Equations:: L'Hopital's rule f'(n) / g'(n)

Hello,This is actually a piece of a little bigger problem (convergence of a series) - you can see the ratio test ak+1 / ak
That's why the (n) and (n+1) terms

I have lim n->∞ of (n√n) / (n+1)√(n+1) ∞/∞

I have tried L'Hopitals rule (requiring multiple times) and I am not seeing an end.

derivative of n^3/2/ (n(n+1)^1/2 + (n+1)^1/2)

lim 3/2 n^1/2 / (1)( n+1)^1/2 + (n)1/2(n+1)^(-1/2) + (1/2 (n+1)^(-1/2) )

form is still ∞/∞

Looking ahead ... it looks as if continuing to apply L'Hopitals rule because of the fractional exponents I will be stuck in a repetitive always getting ∞/∞

The answer (just the answer) was provided to this problem, the series does converge (within a bound) so I know this limit is not ∞/∞

Am I on the right track with L'Hopitals Rule or am I overlooking some simplification to make things easier?

Thanks
Sparky_

Yes, you are overlooking something simpler. Think about the product of two sequences: $\frac{n}{n+1}$ and $\sqrt{\frac{n}{n+1}}$.

 

[Orodruin]

Homework Statement:: Find limit an n-> infinity (n*sqrt(n)) / (n+1)sqrt(n+1))
Relevant Equations:: L'Hopital's rule f'(n) / g'(n)

Am I on the right track with L'Hopitals Rule or am I overlooking some simplification to make things easier?

Why don’t you just rewrite the expression? Divide both numerator and denominator by $n\sqrt n$.

 

[Sparky_]

Yes, you are overlooking something simpler. Think about the product of two sequences: $\frac{n}{n+1}$ and $\sqrt{\frac{n}{n+1}}$.

PeroK - the overall Problem is find the interval of convergence or what values of x ... for x^n / (n sqrt(n) 10^n) I was attempting to use ration test to get to x="10" does this change your reply? Meaning the n and n+1 terms are a result of me implementing the ratio test but I now have to do the limit

 

[Sparky_]

Why don’t you just rewrite the expression? Divide both numerator and denominator by $n\sqrt n$.

OroDruin ...

is this the right direction: lim n-> ∞ n sqrt (n) / (n(sqrt(n+1) + sqrt (n+1))

1 / ( n sqrt(n+1) / n(sqrt(n) + sqrt (n+1) / nsqrt(n) )

= 1 / ( sqrt (n+1) / sqrt (n) + sqrt (n+1) / nsqrt(n))

I see the right most term will go to 0 because of the n term in the denominator

is my next step L'Hoptal's rule?

 

[PeroK]

PeroK - the overall Problem is find the interval of convergence or what values of x ... for x^n / (n sqrt(n) 10^n) I was attempting to use ration test to get to x="10" does this change your reply? Meaning the n and n+1 terms are a result of me implementing the ratio test but I now have to do the limit

That doesn't change my answer. You just have to put things together.

 

[Orodruin]

Consider the following relation:
$$
\frac{n}{n+1} = \frac{1}{1 + 1/n}
$$

 

[Sparky_]

Consider the following relation:
$$
\frac{n}{n+1} = \frac{1}{1 + 1/n}
$$

I think I am with you ... That's how I got 1 / (sqrt(n+1) / Sqrt (n) + sqrt(n+1) / nsqrt(n))

right?

 

[Sparky_]

That doesn't change my answer. You just have to put things together.

I have not caught up to you (yet) ... I should be looking toward the product of (n / n+1) and sqrt (n/n+1)

(n / (n+1) ) ^3/2 ??

 

[Sparky_]

That doesn't change my answer. You just have to put things together.

Root Test?

Lim n-> ∞ x / n^1/n n^1/2n (10) = x/10

 

[PeroK]

I have not caught up to you (yet) ... I should be looking toward the product of (n / n+1) and sqrt (n/n+1)

(n / (n+1) ) ^3/2??

Can you not see (or prove) that
$$\lim_{n \rightarrow \infty} \frac{n}{n+1} = 1$$
If you can't see that, how have you progressed to the radius of convergence for power series?

 

[Mark44]

I think I am with you ... That's how I got 1 / (sqrt(n+1) / Sqrt (n) + sqrt(n+1) / nsqrt(n))

right?

This is awful. Is this anywhere close to what you meant?
$$\frac 1 {\frac{\sqrt{n+1}}{\sqrt n}} + \frac{\sqrt{n+1}}{n\sqrt n}$$
If you right-click on the expression above, you can see the LaTeX script I wrote. There's also a link to our tutorial in the lower-left corner of the input pane.

 

[Mark44]

I have tried L'Hopitals rule (requiring multiple times) and I am not seeing an end.

L'Hopital's Rule isn't always useful, as you found out. For limits involving radicals, it's often better to manipulate the expression algebraically until you get something you can apply the limit to.

 

[Sparky_]

Can you not see (or prove) that
$$\lim_{n \rightarrow \infty} \frac{n}{n+1} = 1$$
If you can't see that, how have you progressed to the radius of convergence for power series?

$$\lim_ { n\rightarrow \infty } \frac {n} {n + 1}$$
$$ = \lim_ { n\rightarrow \infty }\frac {\frac {n} {n} } {\frac {n} {n} + \frac{1} {n}}$$
$$ = \frac {1} {1 + 0}$$

I have not practiced (used) limits in about 40 years.
The background to this question is I was asked by a students parent if I could help with her calculus (having to be home and do the remote learning)

I told the parents yes without asking the topic, I am comfortable with derivatives, integrals differential equations and so forth (I do mathematical modeling as part of my job) I knew when they said series was the topic I was in trouble. Same time frame since doing series - just never really used in my graduate work nor work, only really touched both limits and series in my undergrad.

I was able to solve it without much effort once I found the root test I was stuck trying to solve it in a painful manner.

 

[Sparky_]

This is awful. Is this anywhere close to what you meant?
$$\frac 1 {\frac{\sqrt{n+1}}{\sqrt n}} + \frac{\sqrt{n+1}}{n\sqrt n}$$
If you right-click on the expression above, you can see the LaTeX script I wrote. There's also a link to our tutorial in the lower-left corner of the input pane.

ThanksI have reviewed the tutorial and practiced some LaTex this morning (see my reply to PeroK)

 

[Bilbo B]

Try substituting n/n+1 with some other variable and from that get a new limit for new variable.

 

[archaic]

Since you have already found an answer:
$$\frac{n\sqrt n}{(n+1)\sqrt{n+1}}=\frac{n}{n+1}\frac{\sqrt n}{\sqrt{n+1}}=\frac{n}{n+1}\sqrt{\frac{n}{n+1}}=\left(\frac{n}{n+1}\right)^\frac{3}{2}$$
$$\frac{n\sqrt n}{(n+1)\sqrt{n+1}}=\frac{n^\frac{3}{2}}{(n+1)^\frac{3}{2}}=\left(\frac{n}{n+1}\right)^\frac{3}{2}$$
$$\frac{n\sqrt n}{(n+1)\sqrt{n+1}}\underset{\infty}{\sim}\frac{n\sqrt n}{n\sqrt{n}}=1$$
$n+1\underset{\infty}{\sim}n$ because $\lim_{n\to\infty}\frac{n}{n+1}=1$.

 

[Mark44]

Try substituting n/n+1 with some other variable and from that get a new limit for new variable.

n/n+1 is $\frac n n + 1$, probably not what you intended. If you write fractions inline in text, use parentheses when the numerator or denominator contains multiple terms. IOW, for what you wrote -- n/(n+1).