Series; convergence, divergence

In summary: So the series is alternating, and the terms go to zero.In summary, the first series is convergent, the second series is conditionally convergent, and the third series is convergent. The first series is convergent by the ratio test, the second series is convergent by the alternating series test, and the third series is convergent because the limit of the nth term goes to zero.
  • #1
jnimagine
178
0
series; convergence, divergence...

Homework Statement


1. sum(infinity,n=1) n!/1.3.5...(2n-1)
2. sum(infinity, n=1) (-1)^n arcsin(-1/n)
3. sum(infinity, n=0) arcsin(1/n^2) / arctan(1/n^2)

The Attempt at a Solution


1. i used the ratio test and then i ended up with lim((n+1)(2n-1)/2n+1)) and when i solve the limit i get infinity...
but it's supposed to be convergent... so I'm obviously doing something wrong... :(
2. by alternating series test, it converges
but it's suppsoed to be conditionally convergent, so i tried using the limit test...
how do you approch this problem?? I tried using l'hospital's rule and stuff... but i ended
up with a mess lol
3. in general, when there's inverse trig functions how do u work it out?
 
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  • #2


Actually for the first question [tex]\sum^{\infty}_{n=1} \frac{n!}{1.3.5...(2n-1)}[/tex] you made a mistake while simplifying:

[tex]\frac{s_{n+1}}{s_{n}} = \frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)}/\frac{n!}{1 \times 3 \times 5 \times ... \times (2n-1)}[/tex]

[tex]\frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)} \times \frac{1 \times 3 \times 5 \times ... \times (2n-1)}{n!}[/tex]


(n+1)!/n! = n+1 (which goes on top)

1x3x5x...x(2n+1) is exactly the same as 1x3x5x...x(2n-1)x(2n+1) therefore they cancel out and you are left with (2n+1) on the denomenator.

[tex]= \frac{n+1}{2n+1}[/tex]

So, [tex]lim_{n\rightarrow \infty}\frac{S_{n+1}}{S_{n}} = \frac{1}{2} < 1[/tex]

Therefore the series is...
 
  • #3


roam said:
Actually for the first question [tex]\sum^{\infty}_{n=1} \frac{n!}{1.3.5...(2n-1)}[/tex] you made a mistake while simplifying:

[tex]\frac{s_{n+1}}{s_{n}} = \frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)}/\frac{n!}{1 \times 3 \times 5 \times ... \times (2n-1)}[/tex]

[tex]\frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)} \times \frac{1 \times 3 \times 5 \times ... \times (2n-1)}{n!}[/tex]


(n+1)!/n! = n+1 (which goes on top)

1x3x5x...x(2n+1) is exactly the same as 1x3x5x...x(2n-1)x(2n+1) therefore they cancel out and you are left with (2n+1) on the denomenator.

[tex]= \frac{n+1}{2n+1}[/tex]

So, [tex]lim_{n\rightarrow \infty}\frac{S_{n+1}}{S_{n}} = \frac{1}{2} < 1[/tex]

Therefore the series is...

ohhhh i get it now~~ thank you! ^^

can anyone help me on inverse trig series that I've written above?? :S
also... i think i made some mistake again in using the ratio test for this question:
sum(infinity, n=1) (-1)^n((5^2n)/n!) i end up with lim(25/n+1)... and it's supposed to be convergent absolutely... so... = (
 
  • #4


jnimagine said:
ohhhh i get it now~~ thank you! ^^

can anyone help me on inverse trig series that I've written above?? :S
also... i think i made some mistake again in using the ratio test for this question:
sum(infinity, n=1) (-1)^n((5^2n)/n!) i end up with lim(25/n+1)... and it's supposed to be convergent absolutely... so... = (

What's the limit of 25/(n+1) again? Note the parentheses, it looks pretty misleading if you leave them off. For the second one try the alternating series test. For the last one your first test for convergence should be to check that the limit of the nth term goes to zero. If it doesn't, it can't possibly converge.
 
  • #5


For the other one I think you must apply the alternating series test.

You can show using calculus (calculate a derivative) that |arcsin(-1/n)| is a decreasing function of n with limit |arcsin(0)| = 0.
 

1. What is series convergence?

Series convergence refers to the behavior of an infinite sequence of numbers when the terms of the sequence are added together. If the sum of the terms approaches a finite number as the number of terms increases, the series is said to converge.

2. How is series convergence tested?

The most commonly used tests for series convergence are the integral test, comparison test, and limit comparison test. These tests involve evaluating the behavior of the series in relation to other known convergent or divergent series.

3. What is series divergence?

If the sum of the terms in an infinite sequence of numbers does not approach a finite number as the number of terms increases, the series is said to diverge. This means that the series does not have a definite sum and can potentially increase or decrease without bound.

4. Can a series converge and diverge at the same time?

No, a series can only either converge or diverge, not both. However, some series may exhibit conditional convergence, where the series converges under certain conditions and diverges under others. In such cases, the series is said to be conditionally convergent.

5. What are some real-life applications of series convergence and divergence?

Series convergence and divergence have many practical applications in fields such as physics, engineering, and finance. For example, in physics, series convergence is used to solve problems involving infinite sums, while in finance, it is used to calculate compound interest. Additionally, understanding series convergence and divergence can help predict the behavior of systems in various industries.

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