Converge absolutely

Apr 2009
83
0
infinity



n=1


for absolute convergence, I tried to compare with n^8/n^9 but I couldn't tell whether 1/n is smaller than the series. Is there another method I can test for absolute convergence? Also, on what basis can I conclude it does not converge absolutely (in other words, if a test for absolute convergence leads to a conclusion of divergence, is it entirely divergent or can it still conditionally converge?), and move on to test for conditional convergence?
 
Aug 2007
3,171
860
USA
Alternating series don't need much to be conditionally convergent. Those terms need to be decreasing and heading for zero. Speed isn't really an issue.

Some of these take a little finesse. Play with it a litte.

\(\displaystyle \frac{4n^{8}+4}{8n^{9}+2} >
\frac{4n^{8}+4}{8n^{9}+8} = \frac{1}{2}\cdot
\frac{n^{8}+1}{n^{9}+1}\)

See if that leads anywhere. If not, try something else.

Are you sure long division won't help you?

Really, work with it and try a few things.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
infinity



n=1


for absolute convergence, I tried to compare with n^8/n^9 but I couldn't tell whether 1/n is smaller than the series. Is there another method I can test for absolute convergence? Also, on what basis can I conclude it does not converge absolutely (in other words, if a test for absolute convergence leads to a conclusion of divergence, is it entirely divergent or can it still conditionally converge?), and move on to test for conditional convergence?
\(\displaystyle \sum_{n = 1}^{\infty}\frac{(-1)^n(4n^8 + 4)}{8n^9 + 2}\).

This series is absolutely convergent if \(\displaystyle \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| < 1\).

Don't try using the ratio test - all this time wasted...


So \(\displaystyle \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{(-1)^n[4(n + 1)^8 + 4]}{8(n + 1)^9 + 2}}{\frac{(-1)^n(4n^8 + 4)}{8n^9 + 2}}\right|\)

\(\displaystyle = \lim_{n \to \infty}\frac{\frac{4(n + 1)^8 + 4}{8(n + 1)^9 + 2}}{\frac{4n^8 + 4}{8n^9 + 2}}\)

\(\displaystyle = \lim_{n \to \infty}\frac{[4(n + 1)^8 + 4](8n^9 + 2)}{(4n^8 + 4)[8(n + 1)^9 + 2]}\)

\(\displaystyle = \lim_{n \to \infty}\frac{[(n + 1)^8 + 1](4n^9 + 1)}{(n^8 + 1)[4(n + 1)^9 + 1]}\).

\(\displaystyle = \lim_{n \to \infty}\)
\(\displaystyle \frac{(n^8 + 8n^7 + 28n^6 + 56n^5 + 70n^4 + 56n^3 + 28n^2 + 8n + 1 + 1)(4n^9 + 1)}{(n^8 + 1)[4(n^9 + 9n^8 + 36n^7 + 84n^6 + 126n^5 + 126n^4 + 84n^3 + 36n^2 + 9n + 1) + 1]}\)

\(\displaystyle =\lim_{n \to \infty}\)
\(\displaystyle \frac{\left(1 + \frac{8}{n} + \frac{28}{n^2} + \frac{56}{n^3} + \frac{70}{n^4} + \frac{56}{n^5} + \frac{28}{n^6} + \frac{8}{n^7} + \frac{2}{n^8}\right)\left(4 + \frac{1}{n^9}\right)}{\left(1 + \frac{1}{n^8}\right)\left[4\left(1 + \frac{9}{n} + \frac{36}{n^2} + \frac{84}{n^3} + \frac{126}{n^4} + \frac{126}{n^5} + \frac{84}{n^6} + \frac{36}{n^7} + \frac{9}{n^8} + \frac{1}{n^9}\right) + \frac{1}{n^9}\right]}\)

\(\displaystyle = \frac{1 \cdot 4}{1 \cdot 4}\)

\(\displaystyle = 1\).
 
Apr 2009
83
0
i dont seem to be able to get anywhere wit those. the way I was taught, we have to use tests for convergence which include comparison, limit comparison, alternating series test, test for divergence, ratio test, root test.
i found out to use the limit comparison test with b=1/n in which it led to a conclusion of divergence. I then moved onto conditionally convergence by using the alternating series test. I'm not sure how to express how it decreases. My professor told me to take the derivative and i end up with (8x^9+2)(32x^7)-(4x^8+4)(72x^8) in the numerator. it's really only visual though because I end up simplifying that down to something like -ax^16+bx^7-cx^8

@tkhunny wow lol. yeah ratio test doesn't really work with problems like this.
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
i found out to use the limit comparison test with b=1/n in which it led to a conclusion of divergence.
Correct.

By the way, the ratio or root tests only tell when the general term converges geometrically, i.e. exponentially fast to 0. That is to say, they're only helpful for series that converge pretty quickly.
When for instance we're dealing with polynomial sequences, we know at first sight they can't be conclusive.

I then moved onto conditionally convergence by using the alternating series test. I'm not sure how to express how it decreases. My professor told me to take the derivative and i end up with (8x^9+2)(32x^7)-(4x^8+4)(72x^8) in the numerator. it's really only visual though because I end up simplifying that down to something like -ax^16+bx^7-cx^8
What you only need is to say that the sequence \(\displaystyle \frac{4n^8+4}{8n^9+2}\) decreases for \(\displaystyle n\) larger than some \(\displaystyle n_0\) (there are only finitely many terms before \(\displaystyle n_0\) so they don't alter the convergence or divergence). And you can see that the numerator of your derivative goes to \(\displaystyle -\infty\) as \(\displaystyle x\to+\infty\), so that it is negative for \(\displaystyle x\) larger than some \(\displaystyle x_0\). As a conclusion, the sequence also decreases after any rank \(\displaystyle n_0\geq x_0\).
 
  • Like
Reactions: krzyrice