Infinite Series

[pka]

\(\displaystyle \L \sum\limits_{k = 1}^\infty {\frac{{k^{\frac{1}{2}} }}{{k^3 + 1}}} < \sum\limits_{k = 1}^\infty {\frac{{k^{\frac{1}{2}} }}{{k^3 }}} = \sum\limits_{k = 1}^\infty {\frac{1}{{k^{\frac{5}{2}} }}}\)

Well, keep in mind that sometimes you use techniques I haven't learned yet. Yesterday, we went over the Comparison, Ratio, and Root Tests.

That is the danger is sites such as this. We are not mindreaders. We don't know what you have to work with. You must tell us in detail about your problem.

 

[warwick]

We are not mindreaders. We don't know what you have to work with. You must tell us in detail about your problem.

I know. I'm just saying sometimes if I'm not "getting it" it might be because I have yet to learn the technique. See, I wasn't sure what exactly to compare to the original series. I was initially using 1/k^3 not k^(1/2)/k^3.

 

[pka]

The idea above is: making the denominator smaller, dropping the +1, makes the fraction larger. That is then a p-series.

 

[warwick]

\(\displaystyle \L \sum\limits_{k = 1}^\infty {\frac{1}{{[k(k+1)]}^{1/2}\)

p (rho) = lim as tends to infinity [1/(k(k+1))^(1/2)] / [1/((k+1)(k+1)+1)^(1/2)]

= lim as tends to infinity [k^2 +3k +2]^(1/2) / [k^2 +k]^(1/2) = infinity

By Ratio Test, p > 1 so the Series diverges

 

[pka]

\(\displaystyle \sum {\frac{1}{{\left[ {k\left( {k + 1} \right)} \right]^{\frac{1}{2}} }}} > \sum {\frac{1}{{\left[ {\left( {k + 1} \right)^2 } \right]^{\frac{1}{2}} }}} = \sum {\frac{1}{{k + 1}}}\)
Learning to really use the comparsion test can save you a lot of worK!

 

[warwick]

\(\displaystyle \L \sum\limits_{k = 1}^\infty {\frac{1}{1+{k}^{1/2}\)

p (rho) = lim as k tends to infinity [1/1+(k+1)^(1/2)] / [1/1+(k)^(1/2)]

= lim as k tends to infinity [1+k^(1/2)] / [1+(k+1)^(1/2)]

By Ratio Test, p > 1 so the Series diverges

 

[warwick]

Classify each series as absolutely convergent, conditionally convergent, or divergent.

#19

\(\displaystyle \L\:\sum^{\infty}_{k=1}\,{(-1)}^{k+1}\frac{k+2}{k(k+3)} \\)

Alternating Series Test:

[ a-sub n+1 / a-sub n ] < 1, so the series is decreasing.

The limit as k approaches infinity of [k+2]/[k^2 +3k] = (L'H) 1 / [2k+3] = 0

This tells me the series converges, but I did the Limit Comparison/Ratio Test and according to those it converges. The book says it is conditionally convergent.

#21

\(\displaystyle \L\:\sum^{\infty}_{k=1}\,{sin}\frac{k pi}{2} \\)

I'm not really sure how to approach this problem.

#27

\(\displaystyle \L\:\sum^{\infty}_{k=1}\,{(-1)}^{k+1}\frac{k^{2}+1}{k^{3}+2} \\)

Alternating Series Test:

k^(3) + 2 > k^(2) + 1

[k^(2) + 1] / [k^(3) + 2] < 1, so the series is decreasing.

The limit as k approaches infinity of [k^(2) + 1] / [k^(3) + 2] = (L'H) 1 / [2 /6k] = 0

This tells me the series converges, but none of the Tests I did show that it diverges. The book says it is conditionally convergent.