MHB Improper Integral: Definition and Evaluation

[karush]

316 Ratio Test

$\tiny{11.6.(4) } $
$$L=\sum_{n=1}^{\infty}\dfrac{\ln(n+1)}{n+1} $$
using the Ratio Test
$$L=\displaystyle\lim_{n \to \infty}
\left|\dfrac{a_{n+1}}{a_n}\right|
=\lim_{n \to \infty}\dfrac{\ln((n+1)+1)}{((n+1)+1)}
=\lim_{n \to \infty}\dfrac{\ln((n+2)}{n+2}=0$$
thus $L<1$ convergent

Ok I think this is correct but the final limit I did via W|A not sure why it is 0

 

[HallsofIvy]

What you have is only the limit of [math]a_{n+1}[/math]! You haven't done the fraction, [math]\frac{a_{n+1}}{a_n}=[/math][math] \frac{ln(n+1)}{n+1}\frac{n+2}{ln(n+2)}=[/math][math] \left(\frac{n+2}{n+1}\right)\left(\frac{ln(n+1)}{ln(n+2)}\right)[/math].

What are the limits of those fractions?

 

[skeeter]

for $n \ge 1$, $\bigg| \dfrac{a_{n+1}}{a_n} \bigg| = \dfrac{\ln(n+2)}{n+2} \cdot \dfrac{n+1}{\ln(n+1)} = \dfrac{\ln(n+2)}{\ln(n+1)} \cdot \dfrac{n+1}{n+2}$

Methinks you'll find the ratio test to be inconclusive ... did you decide to use the ratio test to determine convergence/divergence of this sum?

I would use the integral test ...

 

[karush]

so its $\dfrac{1}{2}$ which is still $L<1$

 

[skeeter]

so its $\dfrac{1}{2}$ which is still $L<1$

How did you arrive at 1/2 ?

Did you read my response? When is the ratio test inconclusive?

 

[karush]

How did you arrive at 1/2 ?

Did you read my response? When is the ratio test inconclusive?

$$\lim _{x\to \infty \:}\left(\frac{\ln \left(x+2\right)}{\ln \left(x+1\right)}\cdot \frac{x+1}{x+2}\right)=1$$

it's inconclusive

not sure why we were asked to use Ratio test thot that was used with radicals

 

[skeeter]

integral test ...

$\displaystyle \int_1^\infty \ln(x+1) \cdot \dfrac{1}{x+1} \,dx$

$u = \ln(x+1) \implies du = \dfrac{dx}{x+1}$

$\displaystyle \int_{\ln(2)}^\infty u \,du$

$\displaystyle \lim_{b \to \infty} \int_{\ln(2)}^b u \, du$

$\displaystyle \lim_{b \to \infty} \bigg[\dfrac{u^2}{2}\bigg]_{\ln(2)}^b$

$\displaystyle \lim_{b \to \infty} \dfrac{b^2 - \ln^2(2)}{2} = \infty$

series diverges

 

[karush]

why did you b??

 

[skeeter]

why did you b??

correct format for an improper integral