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Use either the comparison test or ratio test to determine which of the following series converge or diverge

k=1∑∞ (3k^2)-3 / (k^5)+1

k=1∑∞ (k^3/4^k)

k=1∑∞ (1/log(k+1))

k=1∑∞ (1/k + log(k))

2007-12-02 04:40:47 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

Let Sum() represent summation 1 to inf. and lim means lim k-->inf
1) Limit Comparison test
Sum(3/k^3) converges because of p-test
3/k^3> (3k^2-3) / ((k^5)+1) because 3k^5-3k^3 < 3k^5+3
Since Sum(3/k^3) converges Sum((3k^2-3) / ((k^5)+1) ) also convegres

2) Ratio Test
lim |(k+1)^3/ (4*4^k) * (4^k/k^3)| = lim (k+1)^3/(4k^3) = 1/4 <1
Since it's <1, Sum(k^3/4^k) converges

3) Comparison test
Sum(1/(k+1)) diverges because of harmonic test
1/log(k+1) > 1/(k+1) because k+1 > log(k+1)
Therefore Sum(1/log(k+1)) diverges because Sum(1/(k+1)) diverges

4)Comparison test
Sum (1/(2k)) diverges because of p-test
1/(k + log(k)) >1/(2k) because k> log(k)
Therefore Sum( 1/(k + log(k))) diverges because Sum (1/(2k)) diverges

2007-12-02 05:11:01 · answer #1 · answered by SOmath4 2 · 0 0

compare with 1/k^3 convergent
Ratio test a_n+1/a_n= (k+1)^3/k^3 *4^k/4^(k+1) ==> 1/4<1 convergent
compare with 1/n 1/log(k+1) > 1/(n+1) divergent
compare with 1/k = 1+log/k ==> divergent
I supposed the denominator is( k+ log k)
if not ,lim ak is not 0 so it is also divergent

2007-12-02 05:02:24 · answer #2 · answered by santmann2002 7 · 0 0

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2016-10-25 07:47:36 · answer #3 · answered by coulanges 4 · 0 0

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