Convergence (Root Rule)

[NaN-Gram]

I tried to use the Root Rule to show convergence. Where can I go from here?

 

[topsquark]

You didn't distribute the summand correctly in the first line.
[math]\sum_1^{ \infty } a_n = \sum_{n = 1}^{ \infty } (-1)^n \dfrac{1 + n}{n^2} = \sum_{ n = 1} \left ( \dfrac{(-1)^n}{n^2} + \dfrac{(-1)^n}{n} \right )[/math]
For the second line you have (rewriting the variables for clairity) [math]\sqrt[n]{a + b} = \sqrt[n]a + \sqrt[n]b[/math]. This is not correct, either.

For the third line you have [math]\dfrac{\sqrt[n]{(-1)^n}}{\sqrt[n]{n^2}} = \dfrac{-1}{n}[/math]. The numerator is only true for even n and the denominator is only true for n = 1.

The ratio test looks like it might be the better choice if you are allowed to use that.

-Dan

 

[Steven G]

I have a question for you. is 3(2+5) = 3*2 + 5 = 6+5=11?

Now that would bother me is that since (2+5) = (5+2) then 3(2+5) should equal 3(5+2) = 3*5 + 2 = 15+2 = 17 and 17\(\displaystyle \neq\)11.

We fix this by distributing the 3 to the 2 and 5. That is 3(2+5) = 3*2 + 3*5= 6 + 15 =21.

 

[pka]

I tried to use the Root Rule to show convergence. Where can I go from here?
View attachment 17255

I really, really dislike this question because it exposes the failures in the ways this is taught.
Here is the question: discuss the convergence of \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}(1 + n)}}{{{n^2}}}} \)
In my experience there is usually a standard theorem associated with this:
If the sequence \(u_n\) is non-negative decreasing with limit zero then the series \(\sum\limits_{n = 1}^\infty {{{( - 1)}^n}{u_n}} \) converges.
Clearly that theorem applies here. Now to answer the question about absolute convergence:
is it true that \(\dfrac{1+n}{n^2}>\dfrac{1}{n}~?\) What does that tell us.