Monday, 2 April 2018

real analysis - Help with convergence tests for series



I have a few questions to ask about series and convergence tests. I have been struggling to study everything fully and if someone can give me advices I will be really thankful.This is what I know so far, and please correct me if I am wrong with anything.








Definition 1.
For the series $\sum_{n=1}^\infty a_n$ we say that it converges if its sequence of partial sums converges. If it doesn't converge, we say it diverges.
Let's say $S_k$ is the partial sum. If $S_k\to \pm \infty$ we say that
the series diverges.
If $S_k\to$ nothing , we say it diverges as well.









Defintion 2.
$\sum_{n=1}^\infty a_n$, if the series converges then the sequence $(a_n)\to 0$ ($n\to \infty$). If $(a_n)\to 0$ then it does not mean that the $\sum_{n=1}^\infty a_n$ converges.








Examples:
$\sum_{n=1}^\infty \frac{1}{n}$ diverges,
$\sum_{n=1}^\infty \frac{1}{n^\alpha}$ (if $ \alpha$ >1 converges, else diverges) etc.



How do we do tests for the convergence of the following series:

a) $\sum_{n=1}^\infty \arctan\frac{\sqrt n -2}{\sqrt n + 2}$

b) $\sum_{n=1}^\infty \ln (1+\frac{1}{n})$




I have a few questions for the absolute convergence.








Definition 3.
For the series $\sum a_n$ we say that it converges absolutely if the series $\sum |a_n|$ converges. If the series converges but doesn't converge absolutely, we say it conditionally converges.









Definition 4.
LEIBNIZ: (alternate series) $\sum (-1)^n \cdot b_n$ , if $b_n \to 0$ , the series converges.
ABELL: $\sum a_n b_n $, if 1) $\sum a_n$ converges
2) sequence $(b_n)$ is decreasing/increasing and bounded, then $\sum a_n b_n $ converges.







How do we check the absolute and normal convergence of the following series:

a) $\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n^2+1}}\cdot \sin \frac{n\pi}{3}$
b) $\sum_{n=1}^\infty (-1)^n \cdot \tan \frac{3}{\sqrt[4]{n}}$

Thanks.


Answer



Hint for a: I believe that $lim_{n \rightarrow \infty}arctan(.)=arctan(lim_{n \rightarrow \infty}( \frac{\sqrt n -2}{\sqrt n + 2}))=\frac{\pi}{4}.$ Therefore, you can deduce a diverges.




Hint for b: The product of positive real numbers $\prod_{n=1}^{\infty}a_n$ converges iff the sum $\sum_{n=1}^{\infty} log(a_n)$ converges.



If you now look at the sequence of partial products for $\prod_{n=1}^{M}\frac{n+1}{n}=(M+1)$, which tends to infinity as M goes to infinity.


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