Is the Series Absolutely Convergent, Conditionally Convergent, or Divergent?

[ineedhelpnow]

determine whether the series is absolutely convergent, conditionally convergent, or divergent.

$\sum_{n=1}^{\infty} (\frac{n^2+2}{3n^2+2})^n$

how to do this problem??

im thinking root test. but how would i tell if its convergent or conditionally convergent?

 

[ineedhelpnow]

the way I am doing it, it converges to 1/3 but how would i tell if it conditionally converges?

 

[MarkFL]

How could it converge to 1/3 when the first term is 3/5?

 

[ineedhelpnow]

what do you mean 3/5?

 

[MarkFL]

what do you mean 3/5?

Evaluate the summand for $n=1$:

$$\left(\frac{1^2+2}{3\cdot1^2+2}\right)^1=\frac{3}{5}$$

 

[ineedhelpnow]

$\sqrt[n]{|a_n|}=\lim_{{n}\to{\infty}}\frac{n^2+2}{3n^2+2}=\lim_{{n}\to{\infty}}\frac{\frac{n^2}{n^2}+\frac{2}{n^2}}{\frac{3n^2}{n^2}+ \frac{2}{n^2}}= \lim_{{n}\to{\infty}}\frac{1+ \frac{2}{n^2}}{3+\frac{2}{n^2}}=\frac{1+0}{3+0}=1/3$

here's what i did.

how else would i determine?

 

[MarkFL]

Okay, what you mean is:

$$C=\limsup_{n\to\infty}\sqrt[n]{\left(\frac{n^2+2}{3n^2+2}\right)^n}=\frac{1}{3}$$

In this case, $C<1$, and so the series converges absolutely.

 

[ineedhelpnow]

how do i do it the right way?

 

[MarkFL]

how do i do it the right way?

You have already demonstrated that:

$$L=\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\frac{1}{3}<1$$

and so the series is absolutely convergent.

 

[ineedhelpnow]

so what was my mistake from the beginning? the 3/5 thing?

 

[MarkFL]

so what was my mistake from the beginning? the 3/5 thing?

When you said:

the way I am doing it, it converges to 1/3 but how would i tell if it conditionally converges?

I naturally interpreted this to mean "it" was the series itself.

 

[ineedhelpnow]

oh i see