Absolutely convergent, Conditionally convergent, or divergent?
Σ → ∞ 1) Since {3n/(7n^2 + 1)} is a sequence which decreases to 0, the series converges by the Alternating Series Test. Details for decreasing claim: (d/dn) 3n/(7n^2 + 1) = (3 - 21n^2)/(7n^2 + 1)^2 < 0 for all n > 0. However, since Σ(n = 1 to ∞) 3n/(7n^2 + 1) diverges (upon limit comparison to Σ(n = 1 to ∞) n/n^2 = Σ(n = 1 to ∞) 1/n, which is the divergent harmonic series), the given series only converges conditionally. ---------- 2) Since Σ(n = 1 to ∞) sqrt(n-1)/(n^2 - 6) converges (upon limit comparison to Σ(n = 1 to ∞) n^(1/2)/n^2 = Σ(n = 1 to ∞) 1/n^(3/2), a convergent p-series), we conclude that the original series is absolutely convergent. I hope this helps!
Absolutely convergent,conditionally convergent, divergent?
a) Using the Ratio Test: r = lim(n→∞) |[3^(n+1)/n!] / [3^n/(n-1)!] ..= lim(n→∞) 3/n ..= 0. Since r = 0 < 1, this series is absolutely convergent. ---- b) Since n is an integer, note that sin(nπ/2) = 0 if n is even, and (-1)^k if n is odd (n = 2k+1 for some integer k). So, we can rewrite this series as Σ(k = 0 to ∞) (-1)^k/(2k+1). This converges by the Alternating Series Test, because {1/(2k+1)} is a decreasing sequence which converges to 0. However, the convergence is conditional, because Σ(k = 0 to ∞) 1/(2k+1) diverges upon limit comparison with the divergent harmonic series: lim(k→∞) (1/(2k+1))/(1/k) = lim(k→∞) k/(2k+1) = 1/2, which is nonzero and finite. ---- c) Using the Ratio Test: r = lim(n→∞) |a_(n+1) / a_n| ..= lim(n→∞) |[((3n-1)/(7n-2)) a_n] / a_n| ..= lim(n→∞) (3n-1)/(7n-2) ..= 3/7. Since r = 3/7 < 1, this series converges (absolutely). ---- I hope this helps!
Absolute convergent, conditionally convergent, or divergent?
First, this converges by the Alternating Series Test, because {n/√(n^3 + 2)} is a decreasing sequence which converges to 0. To show decreasing behavior: Let g(x) = x/√(x^3 + 2) = x(x^3 + 2)^(-1/2) ==> g'(x) = 1(x^3 + 2)^(-1/2) + x * (-1/2)(x^3 + 2)^(-3/2) * 3x^2 ..............= (1/2)(x^3 + 2)^(-3/2) [2(x^3 + 2) - 3x^3] ..............= (4 - x^3) / [2(x^3 + 2)^(3/2)]. Since g'(x) < 0 for x > 4^(1/3), we see that g is decreasing for x > 4^(1/3). ==> {n/√(n^3 + 2)} is decreasing for n = 2, 3, ... **All that matters is that the sequence decreases for sufficiently large n... --------------------- However, this series is only conditionally convergent, because Σ(n = 1 to ∞) n/√(n^3 + 2) diverges. This is immediate from the Limit Comparison Test with Σ(n = 1 to ∞) n/√(n^3 + 0) = Σ(n = 1 to ∞) 1/n^(1/2), which is a divergent p-series, since lim(n→∞) [n/√(n^3 + 2)] / (1/n^(1/2)) = 1. I hope this helps!
Absolute/conditional convergence and divergence?
(a) is conditionally convergent. We're basically looking at sqrt(n)/n, which is the same as 1/sqrt(n), or 1/(n^(1/2)). We know by the p-test that this does not converge absolutely, but the limit of the corresponding sequence is 0, so the alternating term [(-1)^(n-1)] allows the series to converge conditionally. (b) is absolutely convergent because the numerator is bounded and the denominator is not. (c) is divergent because n! grows faster than 10^n.
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.?
a) This series converges by the Alternating Series Test. Clearly {n/(n^3 + 8)^(1/2)} converges to 0 (by L'Hopital's Rule). To show that {n/(n^3 + 8)^(1/2)} is decreasing: Let g(x) = x/(x^3 + 8)^(1/2). So, g'(x) = [(x^3 + 8)^(1/2) - x * 3x^2 (x^3 + 8)^(-1/2)] / (x^3 + 8) .............= [(x^3 + 8) - x * 3x^2] / (x^3 + 8)^(3/2) .............= (8 - 2x^3) / (x^3 + 8)^(3/2) Since g is decreasing for x > 4^(1/3), we see that g is decreasing for all x > 4^(1/3). Hence, {n/(n^3 + 8)^(1/2)} is decreasing for all n > 1 (which is fine; all we need is that the sequence is decreasing for sufficiently large n). --- However, this series only converges conditionally, because Σ(n=1 to ∞) n/(n^3 + 8)^(1/2) is divergent. To show this, note that n/(n^3 + 8)^(1/2) ≥ n/(n^3 + 8n^3)^(1/2) = 1/(3n) for all n ≥ 1. Since Σ(n=1 to ∞) 1/(3n) is a constant multiple of the divergent harmonic series, the claim follows from the Comparison Test. ---------------------------- b) Since this series has no negative terms convergence and absolute convergence are synonymous. Note that 15n/[(n+1) 5^(2n+1)] < 15n/[n 5^(2n+1)] = 15/5^(2n+1) = 3/25^n for all n. Since Σ(n=1 to ∞) 3 / 25^n is a convergent geometric series, the series in question also converges by the Comparison Test. --------------------------- c) This series converges by the Alternating Series Test, because {1/ln(5n)} is a decreasing sequence which converges to 0. However, this convergence is only conditional, because Σ(n=2 to ∞) 1/ln(5n) diverges. To show this, note that 1/ln(5n) > 1/(5n) for all n. Since Σ(n=2 to ∞) 1/(5n) is a constant multiple of the divergent harmonic series, the claim follows from the Comparison Test. -------------------------- d) Use the root test. r = lim(n→∞) |(-3n/(n+1))^(4n)|^(1/n) ..= lim(n→∞) (-3n/(n+1))^4 ..= (-3)^4 ..= 81. Since r = 81 > 1, this series diverges. ---------------------- I hope this helps!
Determine whether the series is conditionally convergent, absolutely convergent or divergent?
It is divergent because the general term n/(7lnn) is not approaching 0, as n approaches infinity. lim n/(7lnn) -> infinity n->infinity
When do we use the absolute convergence and conditional convergence tests?
If a series is convergent absolutely then it will always be convergent conditionally. But the reverse is not true. A series may converge conditionally but not absolutely.So here's how to know.Take the absolute value of the series, which means ignore the negative alternating sign. Apply a test. If it was convergent then you're done.If divergent then apply AST to the alternating series. If convergent, then the series is conditionally convergent.Otherwise it's always divergent.
What is the radius of convergence of a conditionally convergent power series?
Power series are always absolutely convergent INSIDE their circle of convergence. ON the circle they might converge at some points and diverge at others. It is only on the circle that conditional convergence is possibly an issue.
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.?
This converges. You can use the root test or the ratio test: Ratio test: Look at [ (k+1)*(2/3)^(k+1)] / [ k*(2/3)^k] = [(k+1)/k]*(2/3) = ( 1 + 1/k )*(2/3) As k --> infinity, 1/k --> 0, so we get 2/3. Since this is < 1, it converges by the ratio test. Root test: Look at [k*(2/3)^k]^(1/k) = k^(1/k) * (2/3). As k--> infinity, k^(1/k) goes to 1. * So the ratio test gives 2/3 which is < 1, so it converges. * This is a well known limit. If you're nor familiar with it, you can prove it as follows: let y = k^(1/k), then ln y = (1/k)*ln k = ln(k) / k. As k--> infinity, this goes to infinity/infinity, so use L'Hopital's Rule: it becomes lim [1/k]/ 1 , and as k-->infinity, this goes to 0. So we have that ln y is going to zero, which means that y itself is going to 1. Thus lim k^(1/k) as k--> infinity is 1.
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.?
Take out the common factors and you get: (2/5)( 1 + (6/8)( 1 + (10/11)( 1 + ... assuming the common factor ratios will continue approaching 1 the series will be divergent edit: Actually it's even better than that, if the apparent trend of adding 4 to the numerator and 3 to the denominator for each new factor continues, all of the following factors will be greater than 1 and the terms will start to actually grow. edit2: I'm not sure what the other answer is trying to say but a decreasing series can be divergent as long as it's decreasing slower or at the same rate as the harmonic series (1+1/2+1/3+1/4+...).