convergence of series with square root in the denominator

[sambellamy]

I am being asked to determine whether ∑_n=4^∞ 1/(n-3)^1/2 is convergent. I am not sure if I am doing this correctly.

I decided to use the ratio test:

a_n=1/(n-3)^1/2 a_n+1=1/(n-2)^1/2

I got | (n-3)^1/2 / (n-2)^1/2 |, and then dividing everything by n:

| (1-3/n)^1/2 / (1-2/n)^1/2 |

It seems the numerator will be larger... is the limit 2/3?

I do not know what to do from there. Using L'Hospital's rule seems like it would continue with the ∞/∞ form and not be helpful. Any hints?

 

[Ishuda]

I am being asked to determine whether ∑_n=4^∞ 1/(n-3)^1/2 is convergent. I am not sure if I am doing this correctly.

I decided to use the ratio test:

a_n=1/(n-3)^1/2 a_n+1=1/(n-2)^1/2

I got | (n-3)^1/2 / (n-2)^1/2 |, and then dividing everything by n:

| (1-3/n)^1/2 / (1-2/n)^1/2 |

It seems the numerator will be larger... is the limit 2/3?

I do not know what to do from there. Using L'Hospital's rule seems like it would continue with the ∞/∞ form and not be helpful. Any hints?

In this particular case the ratio test doesn't help because
$\displaystyle lim_{n -> \infty} \frac{a_{n+1}}{a_n} = lim_{n ->\infty} \frac {(n-3)^{\frac{1}{2}}}{(n-2)^{\frac{1}{2}}} =lim_{n ->\infty} [\frac {n-3}{n-2}]^{\frac{1}{2}} = 1^{\frac{1}{2}} = 1$

If you make the substitution m = n-3 and use the comparison test against the harmonic series, you might get further.

 

[pka]

I am being asked to determine whether ∑_n=4^∞ 1/(n-3)^1/2 is convergent.

$\displaystyle n>4,~~\dfrac{1}{\sqrt{n-3}}>\dfrac{1}{\sqrt{n}}>\dfrac{1}{n}$

 

[sambellamy]

I can see that if i just consider

lim_n→∞ 1/(n-3)^1/2

that as the denominator grows, the value will approach 1/∞ which equals zero in the world of limits. is this assessment enough? can I just make this logical leap with the standard test for divergence?

I also have not learned the m substitution method, and I think I'll stay away from it, at least for homework purposes.

 

[pka]

I can see that if i just consider

lim_n→∞ 1/(n-3)^1/2

that as the denominator grows, the value will approach 1/∞ which equals zero in the world of limits. is this assessment enough? can I just make this logical leap with the standard test for divergence?

I also have not learned the m substitution method, and I think I'll stay away from it, at least for homework purposes.

Use simple comparison. $\displaystyle \sum\limits_{n = 5}^\infty {\frac{1}{n}}$ diverges.

 

[sambellamy]

Of course - an easy comparison to make. Thank you!