# Power Series and Integral test

so I'm studying for my exams and there are a few questions that I don't completely understand. I need help with questions (b) and (d). For (b), I ended up getting diverging, because the limit is infinity?. Also I have no clue how to do (d) So I also spent all night trying to learn power series and I need help with (d), (e) and (f). I ended up obtaining the right answer for (d), but I'm not sure if my method of working it out is correct. So basically what I did first was do the ratio test and I ended up getting |x^2|limit k->infinity (log(1+k))/log(2+k). Am I suppose to use L'hopital's rule to find the limit? Because I used it and the interval ended up being from -1infinity (k^k)/(l+1)^k ,what do I do from there? , Btw the answers to (b) and (d) for the integral test is converging for both of them. And the answers to (d), (e) and (f) for the power series is 1, 2e and 0, respectively.

• Help is very much appreciated :) – George Randall May 9 '13 at 16:46

## 2 Answers

Under "Integral Test":

A useful result all students should work out at least once in their "calculus lives" is $\int_1^{\infty} \frac{1}{x^p} \ dx \ ,$ to see for what values of $p$ this integral converges or diverges (you may have already seen this when you covered Type I improper integrals). It will help in spotting which "p-series" $\Sigma_{n=1}^{\infty} \frac{1}{n^p}$ converge.

In your set, you'll need to use u-substitution in examining $\int_0^{\infty} \frac{1}{(n+1)^{\gamma}} \ dx \ ,$ and partial fraction decomposition for $\int_0^{\infty} \frac{1}{(x+1) \ \cdot \ (x+2)} \ dx \$ (or use the hint). (And having re-read your last sentences, yes, (b) and (d) converge.) $$\\$$

Under "Power Series" :

I believe you are correct for (d), (e) , and (f). The Ratio Test for (d) produces

$$\lim_{k \rightarrow \infty} | \ \frac{x^{2k+2}}{x^{2k}} \ \cdot \ \frac{\log (k+1)}{\log (k+2)} \ | \ = \ \lim_{k \rightarrow \infty} | \ x^2 \ \cdot \ \frac{\log (k+1)}{\log (k+2)} \ | ,$$

and I don't think you need a lot of justification to declare that the limit for the ratio of logarithms is $1$ . (The ratio is equivalent to $\log_{k + 2} (k+1)$ ).

As for (e) and (f), these both hinge on dealing with $k^k$ in some manner. For (e), the ratio includes the factors

$$\frac{(k+1)! \cdot k^k}{k! \cdot (k+1)^{(k+1)}} \ ,$$

which reduce to

$$(k+1) \ \cdot \ \frac{ k^k}{ (k+1)^{(k+1)}} \ = \ ( \frac{k}{ k+1})^k \ ,$$

for which the limit at infinity can be found by appropriate use of l'Hopital's Rule on "indeterminate powers", or familiarity with the behavior of this function (as many members of this forum handle it).

Here is one example, your first one. When you integrate that square root term, you get 2*SQRT[x+1] and so this integral is divergent (WHY?) Therefore the Series is divergent. Third example, the anti derivative is an arctan, so can you show that this one will be convergent?

First example of Radius of Convergence, with ratio test (check your book for definition!) you can find that interval of convergence is between -3 and 3. So what's the Radius? Now you can try some more...