Maths calculus 2 Essay
Maths calculus 2
1) Find the sum of (1/(n^2+1)^1/2) from n=1 to infinity Hence, 2) Find the sum of 1/(1+n^1/2) from n=1 to infinity. Hence, 3) Find the sum of from n= 1 to ?. Let’s do some estimation: . Hence, . . 4) Find the sum of 1/(n! ) from n = 1 to ?. It may be easy show the following: It can be noticed, that is the sum of the geometric series. The sum of the geometric progression is ; in our case . Thus . Hence: . As the monotone increasing series is confined by the 3, the series converge and has the finite sum.
The value of the sum of this series calls Euler number and denote as ‘e’. e = 2. 7182818284… Finally: . 5) Test the series for convergence or divergence. . The general term of the series is . It is clear that . Now it is clear, that this limit doesn’t exist. The necessary criterion of the divergence says the terms of the series must converge to 0. Hence the necessary criterion of the divergence of the series isn’t satisfied. Hence, this series doesn’t converge. 6) Test the series for convergence or divergence. The general term of the series is . The series is alternatingsign.
The Leibnitz theorem says that the alternatingsign series converges when absolute magnitude of the terms of the series is monotone decreasing and converge to the zero: . In our case , and hence . Besides that . All theorem conditions are satisfied hence the series converges. 7) Test the series for convergence or divergence The general term of the series is . The series is alternatingsign. The Leibnitz theorem says that the alternatingsign series converges when absolute magnitude of the terms of the series is monotone decreasing and converge to the zero: .
In our case , and (as is monotone increasing function and hence is monotone decreasing function) and hence . Besides that (as ). All theorem conditions are satisfied hence the series converges. 8) Test the series for convergence or divergence. The sum of . From n=1 to ?. The series is alternatingsign. The Leibnitz theorem says that the alternatingsign series converges when absolute magnitude of the terms of the series is monotone decreasing and converge to the zero: . In our case , and . Hence, .
Besides that . All theorem conditions are satisfied hence the sum of converges. 9) Test the series for convergence or divergence: The general term of the series is . It is clear that . Now it is clear, that this limit doesn’t exist. The necessary criterion of the divergence says the terms of the series must converge to 0. Hence the necessary criterion of the divergence of our series isn’t satisfied. Hence, this series doesn’t converge. 10) What can you say about the series an in each of the following cases? (a) (b) (c)
(a) Solution: if then At sufficiently great n we have where . Hence, . Lets find the . Discriminant: . As we can conclude that . Hence, . (b) Solution: if then At sufficiently great n we have where . Hence, . Lets find the . Discriminant: . As , we can conclude that . Hence there is no exist such series an, that !!! (c) Solution: if then At sufficiently great n we have where . Hence, . Lets find the . Discriminant: . As , we can conclude that . Hence there is no exist such series an, that !!!
B

Subject: Leibnitz,

University/College: University of Arkansas System

Type of paper: Thesis/Dissertation Chapter

Date: 17 May 2017

Words:

Pages:
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