Series Convergent or Divergent?
For each of the series below select the letter from a to c that best applies and the letter from d to k that best applies
A. The series is absolutely convergent.
B. The series converges, but not absolutely.
C. The series diverges.
D. The alternating series test shows the series converges.
E. The series is a and#92;(pand#92;)-series.
F. The series is a geometric series.
G. We can decide whether this series converges by comparison with a and#92;(pand#92;) series.
H. We can decide whether this series converges by comparison with a geometric series.
I. Partial sums of the series telescope.
J. The terms of the series do not have limit zero.
K. None of the above reasons applies to the convergence or divergence of the series.
1. summation from 2 to INF of 1/(nlog(6 n))
2. summation from 1 to INF of 1/(nsqrt(n))
3.summation from 1 to INF of (cos(npi))/(npi)
4. summation from 1 to INF of ((2n 4)!)/((n!)^2)
5. summation from 1 to INF of (cos^2(npi))/(npi)
6. summation from 1 to INF of (5 sin(n))/(sqrt(n))
Comments
1) Choices C and K.
Note that 1/(n log(6 n)) > 1/((n 6)log(n 6)) for all n > 1.
Since ∫(x = 1 to ∞) dx/((x 6) log(x 6))
= ∫(u = ln 7 to ∞) du/u, letting u = log(x 6) assuming base e
= ln |u| for u = ln 7 to ∞
= ∞ (divergent),
we see that Σ(n = 2 to ∞) 1/((n 6)log(n 6)) diverges by the Integral Test.
==> Σ(n = 2 to ∞) 1/(n log(n 6)) diverges by the Comparison Test.
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2) Choices A and E.
The nth term is 1/n^(3/2). So, this is a p-series with p = 3/2 > 1.
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3) cos(nπ) = (-1)^n for any integer n; so we have an alternating series.
Choices B and D.
This converges by Alternating Series Test, because 1/(nπ) decreases to 0.
The convergence is conditional because Σ(n = 1 to ∞) 1/(nπ) diverges (multiple of harmonic series).
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4) Choices C and J.
r = lim(n→∞) [(2n 6)!/((n 1)!)^2] / [(2n 4)!/(n!)^2]
..= lim(n→∞) (2n 6)(2n 5)/(n 1)^2
..= 4 > 1.
So, this series diverges by the Ratio Test.
However, we can show that the nth term does not converge to 0 by Stirling's Estimate:
lim(n→∞) (2n 4)!/(n!)^2
= lim(n→∞) (2n 4)(2n 3)(2n 2)(2n 1) * (2n)!/(n!)^2
= lim(n→∞) (2n 4)(2n 3)(2n 2)(2n 1) * [(2n/e)^(2n) √(2π*2n)]/[(n/e)^n √(2πn)]^2, by Stirling
= lim(n→∞) (2n 4)(2n 3)(2n 2)(2n 1) * [2^(2n) (n/e)^(2n) * 2√(πn)]/[(n/e)^(2n) (2πn)]
= lim(n→∞) (2n 4)(2n 3)(2n 2)(2n 1) 2^(2n) / √(πn)
= ∞ (by one application of L'Hopital's Rule or otherwise).
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5) Choices C and E.
cos(nπ) = (-1)^n for any integer n; so we have an alternating series.
So, this reduces to Σ(n = 1 to ∞) 1/(nπ), which diverges (multiple of harmonic series).
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6) Choices C and G.
Note that (5 sin n)/sqrt(n) > (5 - 1)/n^(1/2) = 4/n^(1/2), and Σ(n = 1 to ∞) 4/n^(1/2) diverges,
being a multiple of a divergent p-series. So, the original series diverges by the Comparison Test.
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I hope this helps!
1. summation from 2 to INF of 1/(nlog(6 n)) C
2. summation from 1 to INF of 1/(nsqrt(n)) A
3.summation from 1 to INF of (cos(npi))/(npi) C
4. summation from 1 to INF of ((2n 4)!)/((n!)^2) C
5. summation from 1 to INF of (cos^2(npi))/(npi) C
6. summation from 1 to INF of (5 sin(n))/(sqrt(n)) C