# Determine convergence or divergence

• May 17th 2010, 06:02 PM
krzyrice
Determine convergence or divergence
http://mathworld.wolfram.com/images/...dEquation2.gif where ak= (7^n)/(3n^6+4^n)

I tried the alternating series test but I get stuck on determining whether or not it is decreasing (using derivative)

Also this other problem test for absolute convergence, conditionally convergent or divergent
infinity
E (-1)^n (4n^8+4)/(8n^9+2)
n=1

how would i test this for absolute convergence? I tried ratio test and it just becomes a bigger mess
• May 17th 2010, 06:46 PM
tonio
Quote:

Originally Posted by krzyrice
http://mathworld.wolfram.com/images/...dEquation2.gif where ak= (7^n)/(3n^6+4^n)

I tried the alternating series test but I get stuck on determining whether or not it is decreasing (using derivative)

Also this other problem test for absolute convergence, conditionally convergent or divergent
infinity
E (-1)^n (4n^8+4)/(8n^9+2)
n=1

how would i test this for absolute convergence? I tried ratio test and it just becomes a bigger mess

The first sum seems to be $\sum^\infty_{k=1}(-1)^k\frac{7^k}{3k^6+4^k}$ (I assume here that you meant the index to be k and not n...), and if this is so the sum can't converge since

the limit of its general term is not zero: $\frac{7^k}{3k^6+4^k}\geq \frac{7^k}{2\cdot 4^k}\xrightarrow [k\to\infty]{}\infty$ , and with the alternating sign the general term doesn't even diverge to infinity.

About the second one: it converges conditionally since it is a Leibnitz series, but taking its general term's absolute value we get:

$\frac{4n^8+4}{8n^9+2}\geq \frac{4n^8}{16n^9}=\frac{1}{4}\,\frac{1}{n}$ , and by the comparison test with the harmonic series it diverges.

Tonio