Limit Comparison Test for Convergence or Divergence: (n+5)/(n^3-2n+3) and n/n^3

[ProPatto16]

Homework Statement

show whether \sum (n+5)/(n³-2n+3) is convergant of divergant

Homework Equations

limit comparison test, lim a_n/b_n = c where c > 0

The Attempt at a Solution

a_n is given
let b_n = n/n³

so then:

lim (n+5)/(n³-2n+3)
n/n³
= lim (n+5)n³ / (n³ - 2n + 3)n
= lim (n⁴ + 5n³) / (n⁴ - 2n² +3n)
= lim (5n³) / (-2n²+3n)

now as n -> infinity, lim (5n³) / (-2n²+3n) -> -5n/2

and as n -> inifinity, -5n/2 -> -inifinity... therefore divergant? since c < 0??


feel free to tell me i have no idea. these things confuse me so much.

 

[Mark44]

Homework Statement

show whether \sum (n+5)/(n³-2n+3) is convergant of divergant

Homework Equations

limit comparison test, lim a_n/b_n = c where c > 0

The Attempt at a Solution

a_n is given
let b_n = n/n³

so then:

lim (n+5)/(n³-2n+3)
n/n³
= lim (n+5)n³ / (n³ - 2n + 3)n
= lim (n⁴ + 5n³) / (n⁴ - 2n² +3n)
= lim (5n³) / (-2n²+3n)

now as n -> infinity, lim (5n³) / (-2n²+3n) -> -5n/2

and as n -> inifinity, -5n/2 -> -inifinity... therefore divergant? since c < 0??


feel free to tell me i have no idea. these things confuse me so much.

Instead of b_n = n/n³, simplify this to 1/n².

The limit you should be working with is
\lim_{n \to \infty} \frac{\frac{n + 5}{n^3 - 2n + 3}}{\frac{1}{n^2}}

 

[ProPatto16]

that limit simplifies to

lim 5n²/(-2n+3)

so then my previous explantion still stands...

now as n -> infinity, lim (5n²) / (-2n+3) -> -5n/2

and as n -> inifinity, -5n/2 -> -inifinity... therefore divergant? since c < 0??

so i dunno...

i also tried taking the simplified limit and dividing through by n² which gives

lim 5/(-2/n + 3/n²) and i know the limit of 1/n or 1/n²
and all varieties is 0... so then

lim 5/(-2/n + 3/n²) = 5/(0-0) = 0... since that is not >0... then it hasnt worked...

 

[snipez90]

No, that is incorrect. Remember for a rational function f(n), if you let n go to infinity only the terms with the largest exponent on n remain in the numerator and in the denominator.

For instance, lim (5n^2 + 1000n)/(3n^2 + 40n + 500) is simply lim (5n^2)/(3n^2) = 5/3. Intuitively as n grows beyond all bounds, 5n^2 completely dominates 1000n in the numerator, and similarly 40n + 500 is insignificant compared to 3n^2 when n is approaching infinity.

 

[Mark44]

that limit simplifies to

lim 5n²/(-2n+3)
so then my previous explantion still stands...

Check your algebra. For large n, the numerator grows at exactly the same rate as the denominator.

 

[ProPatto16]

yeah i used that method of thinking to reduce the limit down:

so then lim 5n2/(-2n+3) becomes lim 5n/-2

but then i don't know where to go from there..

 

[Mark44]

yeah i used that method of thinking to reduce the limit down:

so then lim 5n2/(-2n+3) becomes lim 5n/-2

but then i don't know where to go from there..

That's still wrong. The limit is a finite constant.

 

[ProPatto16]

mark... yeah i did... it all comes down to a ratio of -2.5n... so the limit still dpeends on n.

if n = 1000000000 then it comes to -2500000004... i can see the effect here... i just can't summarise it in a way that gives me an appropriate answer...

 

[ProPatto16]

the only finite constant here that would make any relevant sense is -5/2...

 

[Mark44]

mark... yeah i did... it all comes down to a ratio of -2.5n... so the limit still dpeends on n.

if n = 1000000000 then it comes to -2500000004... i can see the effect here... i just can't summarise it in a way that gives me an appropriate answer...

No, it does not depend on n.

the only finite constant here that would make any relevant sense is -5/2...

No.

The numerator is (n + 5)/(n^3 - 2n + 3). The denominator is 1/n^2. For large n, the numerator is roughly n/n^3 or 1/n^2.

CHECK YOUR ALGEBRA!

 

[ProPatto16]

(n+5)/(n³-2n+3) is the numerator. for large values of n... this becomes n/n³

the denominator is 1/n²

the numerator can simplify down to 1/n² also.

so then:

lim (n+5)/(n³-2n+3) / 1/n²

becomes lim 1/n² / 1/n²

= 1
since this is greater than 0... both functions converge or diverge... and since 1/n² converges... so does (n+5)/(n³-2n+3) ??

 

[Mark44]

(n+5)/(n³-2n+3) is the numerator. for large values of n... this becomes n/n³

Which is 1/n^2.

the denominator is 1/n²

the numerator can simplify down to 1/n² also.

so then:

lim (n+5)/(n³-2n+3) / 1/n²

becomes lim 1/n² / 1/n²

= 1

YES!

since this is greater than 0... both functions converge or diverge... and since 1/n² converges... so does (n+5)/(n³-2n+3) ??

Yes.

 

[ProPatto16]

thank you. sorry for being frustrating.

 

[Mark44]

Here are the steps in evaluating that limit.
\lim_{n \to \infty} \frac{\frac{n + 5}{n^3 - 2n + 3}}{\frac{1}{n^2}}
=\lim_{n \to \infty} \frac{n + 5}{n^3 - 2n + 3} \frac{n^2}{1}
=\lim_{n \to \infty} \frac{n(1 + 5/n)}{n^3(1 - 2/n^2 + 3/n^3} \frac{n^2}{1}
=\lim_{n \to \infty} \frac{(1 + 5/n)}{(1 - 2/n^2 + 3/n^3} \frac{1}{1}
= 1

I did quite a bit of cancelling after factoring out n, n^3, and so forth. All of the terms with n to some power in the denominator go to zero when n gets large.

 

[ProPatto16]

i see... factorise out the highest exponent that appears in the equation.

thanks a lot:)