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.
For what value(s) of x does the series converge absolutely and conditionally?
a) use the ratio test, and the fact that sqrt( n/(n+1) ) -> 1 as n->infinity. This will give you |x-3|<1, or 2 < x < 4, for absolute convergence. b) the ratio test is inconclusive for |x-3|=1, so let's look specifically at those values of x. For x=4, we have the series Sum( 1 / sqrt(n) ), which diverges (use p-series, or comparison with Integral(1/x^(1/2) ) ) For x=2, we have the alternating series Sum( (-1)^n / sqrt(n) ) {1/sqrt(n)} is a positive decreasing sequence. So by the Leibniz test, this series converge (and conditionally, as we've seen for x=4).
Divergent or convergent series question? (Calculus II)?
oo Σ (-1)^n * lnn * n^-1 n=1 Σ (-1)^n * lnn / n I don't see why the Alternating Series Test wouldn't work. Let's apply it. Let's call b_n lnn/n. lim n-->oo b_n = lim n-->oo lnn / n = indeterminate form. lim n-->oo (1/n)/1 = 0. Now let's check if b_n is a decreasing sequence. ln1/1, ln2/2, ln3/3, ln4/4, ln5/5, ln6/6.. ....0... .34.... .36.... .34......32.... .29.. The series is decreasing for all values of n > 3. We can also take the derivative to prove that indeed the sequence of numbers is decreasing. y = lnn / n y' = (1/n)(1/n) + lnn(-1/n^2) y' = 1/n^2 - lnn/n^2 y' = (1 - lnn) / n^2 0 = 1 - lnn 1 = lnn e = n This is our only critical value. Let's test some points around it to see what's happening to the slope. f '(3) = (1 - ln3 / 3^2 ............ = -.09 / 9 = -.01 f '(2) = (1 - ln2) / 2^2 ........= .3 / 4 = .07. This tells us that the slope is negative for values greater than e. Therefore, the values of the sequence are decreasing after e. By the Alternating Series Test, the series converges. Now, to test for Conditional Convergence, let's take the absolute value of the series. oo Σ | -1)^n * lnn / n | = n=1 Σ lnn / n = Let's use the Integral Test to check if this series converges or not. ∫ lnx / x dx Let, u = lnx du = 1/x dx ∫ u du = u^2/2 = (lnx)^2/2 [1 , oo] (1/2)(ln(oo))^2 - (ln1)^2) = oo - 0 = oo. By the Integral Test, the series diverges. Hence, the series is Conditionally Convergent. Hope this helped.
Find all the values of x such that the given series would converge.?
The presence of powers suggests that we should use the Ratio Test. Notice that, by the laws of exponents: (-1) * 4^n * x^n = (-1 * 4 * x)^n = (-4x)^n, so we have: ∑ (-4x)^n/(√n + 8) (from n=1 to infinity). (It doesn't matter where the series starts as long as it starts at a positive integer.) By the Ratio Test, the series ∑ a_n (from n=1 to infinity) (or any other starting point) converges if: |lim (n-->infinity) (a_n+1/a_n)| < 1, and is inconclusive when: |lim (n-->infinity) (a_n+1/|a_n)| = 1. (We need to test the values of x that cause this separately using other tests.) With a_n = (-4x)^n/(√n + 8), we see that a_n+1 = (-4x)^(n + 1)/[√(n + 1) + 8] and: a_n+1/a_n = {(-4x)^(n + 1)/[√(n + 1) + 8]}/[(-4x)^n/(√n + 8)] = [(-4x)^(n + 1) * (√n + 8)]/{(-4x)^n * [√(n + 1) + 8]} = [(-4x) * (√n + 8)]/[√(n + 1) + 8], by canceling (-4x)^n. (√n + 8)/[√(n + 1) + 8] --> 1 as n --> infinity, so: |lim (n-->infinity) (a_n+1/a_n)| = |-4x| = 4|x|, which is less than 1 when: 4|x| < 1 ==> |x| < 1/4 ==> -1/4 < x < 1/4. (When x is in this range, the series converges ABSOLUTELY.) At the endpoints of this interval (x = -1/4 and x = 1/4), |lim (n-->infinity) (a_n+1/a_n)| = 1, so we need to test these two cases separately. With x = -1/4, the series becomes: ∑ (-4 * -1/4)^n/(√n + 8) (from n=1 to infinity) = ∑ 1/(√n + 8) (from n=1 to infinity), which diverges due to its similarity to the series: ∑ 1/√n (from n=1 to infinity), which is a divergent p-series with p = 1/2 < 1. With x = 1/4, the series becomes: ∑ (-4 * 1/4)^n/(√n + 8) (from n=1 to infinity) = ∑ (-1)^n/(√n + 8) (from n=1 to infinity), which converges by the AST since 1/(√n + 8) --> 0 and a_n = 1/(√n + 8) is a monotonically decreasing sequence. Notice that this series converges conditionally since: ∑ |(-1)^n/(√n + 8)| (from n=1 to infinity) = ∑ 1/(√n + 8) (from n=1 to infinity), is a divergent series (see the first case). Therefore, the series converges for -1/4 < x ≤ 1/4. I hope this helps!
Basic Calculus 2 Converge/Diverge problems?
Checking my homework, the answers are multiple choice, so I'll just be matching them. Explanation is nice but not required, please don't use wolfram alpha, These are the ones that website cannot help me with. Easy ten points, cheers! If you'd like, post your paypal email, I'll gratify you for the help, I know it's a lot! I don't mind sending over 50-100 bucks, you know tutors are expensive... Thanks guys Determine if convergent or divergent please tell me if i'm right/answer 1)Sum 1/(2n-1) n=1 to infinity I answered Divergent 2)sum 8n/n! n=1 to infinity I answered Convergent 3)Sum (-1)^n/(n^5/3) n=1 to infinity I answered convergent 4)sum ((lnn)/(3n+7))^n n=1 to infinity I answered convergent 5)sum (x-1)^n/(3n+4) n=0 to infinity ?????? 6)a1=10, a_n+1 = (n/(n+1))a_n Does this converge or diverge? 7)Find the Maclaurin series for e^-7x They want the Sum structure, I put 7nx^n/n! but I don't think that's right. 8)Sum (x-2)^n n=1 to infinity I put -x-2/x-1 but I keep thinking the 1 has to be positive, please confirm. 9)Sum (x-9)^n n=0 to infinity What's the radius of convergence? 10)Sum (-1)^n/(9n^(1/4)+1) I answered that it converges conditionally, because it alternates signs, right? 11)What's the Taylor series for ln(1+x^2)? the sum formula 12)a_n=(-1)^n (1-(7/n)) does it converge or diverge? what's the limit 13)Write the first four elements of the sequence (1+(1/n))n I think the last n is a subscript, can't figure this out 14)What's the limit n!/9^n*5^n 15)Values of X where it converges sum(x-10)^n n=0 to infinity 16)Find the sum of the series ((1/(7n))+(1/(4n))) n=0 to infinity the n=0 screws me up help! 17)What's the limit (1+8/n)n that n is a subscript (the outside one) I said it diverges.
Why can the limit of a function be zero, but the series still diverge?
Note that even though the terms converge to 0, the series can diverge (i.e. the partial sums could still grow without bound) if the terms converge too slowly to 0. Here's an example using the series from k = 1 to infinity of 1/k, called the harmonic series. Note that 1/k converges to 0. However, 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 +... >= 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 +... = 1 + 1/2 + 2(1/4) + 4(1/8) + 8(1/16) + ... = 1 + 1/2 + 1/2 + 1/2 + 1/2 + ... = infinity So the series from k = 1 to infinity of 1/k diverges to infinity. By the way, it would then follow that the series from k = 1 to infinity of 1/(k^(1/2)) also diverges to infinity, since 1/(k^(1/2)) >= 1/k > 0, for k >= 1. Here's an interesting fact: the series from k = 1 to infinity of 1/k diverges to infinity, and yet if p is even slightly greater than 1, then the series from k = 1 to infinity of 1/(k^p) converges!
Finding radius and interval of convergence of a series?
Use the Ratio Test. Here, the nth term a_n is given by a_n = 2^(n+1)*(x-2)^n/(3n^2). So, r = lim(n→∞) |a_(n+1) / a_n| = lim(n→∞) |[2^(n+2)*(x-2)^(n+1)/(3(n+1)^2)] / [2^(n+1)*(x-2)^n/(3n^2)]| = 2 |x - 2| * lim(n→∞) n^2/(n+1)^2 = 2|x - 2|. So, this series converges (at least) for r = 2 |x - 2| < 1 ==> |x - 2| < 1/2. ==> Thus, the radius of convergence is 1/2. As for the interval of convergence, we next test the endpoints. Since |x - 2| < 1/2 ==> 3/2 < x < 5/2: (i) x = 5/2 ==> Σ 2^(n+1) (1/2)^n / (3n^2) = (2/3) Σ 1/n^2. This is a multiple of a convergent p-series; hence it converges. (ii) x = 3/2 ==> Σ 2^(n+1) (-1/2)^n / (3n^2) = (2/3) Σ (-1)^n/n^2. This absolutely converges by (i). So, the interval of convergence is [3/2, 5/2]. I hope this helps!
Calc help - sequence/series?
You must first find the derivative f ' (x). The function will be decreasing for all values of x where the derivative f ' (x) is < 0. You will get f ' (x) = (57 - x²)/(x² + 7x + 57)² This will be negative if x > sqrt(57). So the answer to (A) is r = sqrt(57). You can make a spreadsheet to calculate values of f(n) for integers n. You will get n f(n) 1 0.015384615 2 0.026666667 3 0.034482759 4 0.03960396 5 0.042735043 6 0.044444444 7 0.04516129 8 0.04519774 9 0.044776119 etc Since 8 > sqrt(57), we know f(n) is going to decrease from n = 8 onward. And we can see from the table of values that f(n) is increasing up to n = 8. So the answer to (B) is s = 8.
Convergence of a Power Series?
a) i agree with ur method of the ratio test as you have reached : x - 2 < 1 .....but the radius of convergence is 1 but if we have it as x/2 < 1 , then x < 2 ( the radius of convergence is 2 ) so our radius of convergence is 1. =========== b) interval : -1 < x - 2 < 1 -1 + 2 < x - 2 + 2 < 1 + 2 1 < x < 3 let's check these endpoints into the series as the following: when x = 1 ∞ ∑ (x - 2)^n / n = (1 - 2)^n / n = (-1)^n / n ( it is a special case for harmonic series converges ) n=0 when x = 3 ∞ ∑ (x - 2)^n / n = (3 - 2)^n / n = (1)^n / n ( take the lim as n---> ∞ 1/n = 1/∞ = 0 converges BUT ) n=0 it diverges by p-series because 1 ≤ 1 <==== diverges. so the interval will be as 1 ≤ x < 3 ================ c) absolute convergent it means if we take the absolute value of the series it becomes positive. for the values x = 3 , 4 , 5 , 6 , 7 , ... for example : (7 - 2)^n / n = 5^n / n <==== that means never absolutely convergent because our n in the denominator its exponent not greater than 1. ============== d) let's start evaluation as x = 1 , 0 , -1 , -2 , -3 , ... (-3 - 2)^n / n ====> -5^n / n ===> (-1)^n * 5^n / n <=== it diverges because of the p-series and if we test it by ratio test as : [ (-1)^(n + 1) * 5^(n + 1) / (n + 1) ] * [ n / (-1)^n * 5^n ] [ (-1)^(n) * -1^1 * 5^(n) * 5^1 / (n + 1) ] * [ n / (-1)^n * 5^n ] -5 * [ n / (n + 1) ] lim -5 * (n / (n + 1) ) ===> large in charge n--> ∞ lim -5 * (n /n) n--> ∞ lim -5 * 1 = -5 ====> I -5 I < 1 ===> 5 < 1 <== not true absolutely divergent n--> ∞ conditionally convergent when x = -1 ========= please e-mail me if u still have a question.