Sum from 0 to infinity of ke^(-k) where k=0,1,2...?
Start off with the formula for the sum of an infinite geometric series. 1/(1 - x) = sum(k=0 to infinity) x^k. Differentiating both sides with respect to x gives: 1/(1 - x)^2 = sum(k=0 to infinity) [k * x^(k - 1)] ==> x/(1 - x)^2 = sum(k=0 to infinity) (k * x^k), by multiplying both sides by x. Then, with x = 1/e: sum(k=0 to infinity) [k * (1/e)^k] = (1/e)/(1 - 1/e)^2 ==> sum(k=0 to infinity) k*e^(-k) = e/(1 - e)^2. I hope this helps!
What is the sum of (cos 1) ^k from k=1 to infinity?
Geometric series formula : [math]a\sum\limits_{k=0}^{\infty}{r^k}=\frac{a}{1-r}\Leftrightarrow|r|<1[/math].Because [math]|\cos(1)|<1[/math], it follows that[math]\sum\limits_{k=1}^{\infty}{\cos^k(1)}=\sum\limits_{k=0}^{\infty}{\cos^k(1)}-\cos^0(1)[/math][math]=\frac{1}{1-\cos(1)}-1=\boxed{\frac{\cos(1)}{1-\cos(1)}}[/math].
HELP w/summation ln ((k+2)/(k+1)) from k=1 to k=infinity?
We have to use any series test to determine whether or not the infinite series converges or diverges. The answer in the back of the book is that it diverges. For me, the root test and ratio test look like they would be too much work. I know the integral test would be too confusing and long for me. Would it be wrong if I used the limit comparison test and said ak = ln((k+2)/(k+1)) and bk = (k+2), which diverges. Then when use L'Hopital's Rule and divide ak/bk, you get the lim as k approaches infinity = 1. And since L = 1, and bk diverges, then ak diverges as well, or would that be wrong?
What is the infinite sum of the series 1/1 + 1/3 + 1/6 + 1/10 + 1/15 +?
Let Tn represents the n-th term in this series.T1 = 1/(1)T2 =,1/(1+2)T3 = 1/(1+2+3)T4 = 1/(1+2+3+4)T5 = 1/(1+2–3+4+5) and so on-> Tn = 1/(1+2+3+……..+n)= 1/ [ n(n+1)/2 ] = 2/n(n+1)Tn = 2 / [ n(n+1) ]= 2 * [ (1/n) - 1/(n+1) ]If you add all the terms you will get >2 * [ 1/1 -1/2 + 1/2 -1/3 + 1/3 - 1/4 +1/4 - 1/5 + 1/5 - 1/6 +…….]Answer = 2
Consider the infinite geometric series: (k=1 to infinity)Σ (cos^2ϴ)^k for 0
Consider the infinite geometric series: (k=1 to infinity)Σ (cos^2ϴ)^k for 0 1. find its sum whenever the series converges a. sin^2ϴcos^2ϴ b. csc^2ϴ c. tan^2ϴ d. sec^2ϴ e. cot^2ϴ 2. find all values of ϴ between 0 and 2pi, at which the series does not converge. a. pi/2, 3pi/2 b. 0, pi/2, pi, 3pi/2 c. 0, pi d. pi/4, 5pi/4 e. pi/4, pi, 7pi/4 please show your steps for solving the problems. Thank you so much for your help! :)
Summation from 1 to infinity of 1/9k^2+3k-2?
The skip from 1/[(3k - 1)(3k + 2)] to (1/3)/(3k - 1) - (1/3)/(3k + 2) came from Partial Fractions. From this step, we can continue to get: sum(n=1 to infinity) 1/(9k^2 + 3k - 2) = sum(n=1 to infinity) [(1/3)/(3k - 1) - (1/3)/(3k + 2)] = 1/3 * sum(n=1 to infinity) 1/(3k - 1) - 1/3 * sum(n=1 to infinity) 1/(3k + 2). To find this sum, consider the partial sum of the first i terms. sum(n=1 to i) 1/(9k^2 + 3k - 2) = 1/3 * sum(n=1 to i) 1/(3k - 1) - 1/3 * sum(n=1 to i) 1/(3k + 2) = 1/3 * [1/2 + 1/5 + 1/8 + ... + 1/(3i - 1)] - 1/3 * [1/5 + 1/8 + 1/11 + ... + 1/(3i - 2)] = 1/3 * [1/2 - 1/(3i - 2)] = 1/6 - 1/[3(3i - 2)]. Letting i --> infinity: sum(n=1 to infinity) 1/(9k^2 + 3k - 2) = lim (i-->infinity) sum(n=1 to i) 1/(9k^2 + 3k - 2) = lim (i-->infinity) {1/6 - 1/[3(3i - 2)]} = 1/6 - 0 = 1/6. I hope this helps!
Summation k=1 to infinity of cos(1)^k?
I'd definitely say you're on the right track. Of course, cos(0) = 1, but I think that cos(1) actually turns out to be an irrational number. And since | cos(1) | < 1, you're right, we could go ahead and treat it like a geometric series. Even though we don't "see" an a, it's safe to go ahead and think of (cos1)^k as being multiplied by 1, and call 1 our a. (If we said that the series' a was 0, that'd mean--since 0 times number is 0--all of the terms would end up being 0, which I don't think is the case for us.) So the sum would be: a/(1-r) = 1/(1-cos1). There you have it. That's all there is to it! I hope this helps.
K=1 and K goes to infinity, what is the sum of series (k-1) /k?
(k-1)/k=1–(1/k) >1/k when k>1The sum of 1/k go to infinity.Thus, the sum of (k-1)/k go to infinity.
Help with sequences and infinite series?
Notice that (-1)^n = 1 for even n and -1 for odd n, so we see that: a_n = (n + 2)/(3n - 1) for even n and -(n + 2)/(3n - 1) for odd n. Since (n + 2)/(3n - 1) --> 1/3 and -(n + 2)/(3n - 1) --> -1/3 as n --> infinity, we see that the even terms converge to 1/3 while the odd terms converge to -1/3; since these two sub-sequences converge to a different limit, the sequence diverges (intuitively, this should make sense since the series keeps oscillating between terms close to -1/3 and 1/3). --- As for the series, notice that we can split the series up into the following two series: ∑ 1/2^k (from k=0 to infinity) - ∑ 1/(k + 1) (from k=0 to infinity). The first series is convergent (it is an infinite geometric series with a common ratio of 1/2), while the second series is the divergent p-series ∑ 1/n (from n=1 to infinity), so since the series can be written as a difference of two series, one of which converges and the other diverges, the series diverges. I hope this helps!