Finding the Number of Terms Needed to Approximate an Infinite Sum within 1/100

[member 508213]

Homework Statement

The problem says approximate the sum to within 1/100 of the value of the infinite sum, and the sum is
1/2-(1x3)/(2x4)+(1x3x5)/(2x4x6) and so on... (Teacher said I can leave answer in summation notation so I just need to find how many terms I need to add together.)

Homework Equations

n/a

The Attempt at a Solution

Since it is an alternating series, I would just need to find the first term that is equal to or less than 1/100 to know how many of the terms I need to add together and the terms are (1x3x...x(2n+1))/(2x4x...(2n+2)). However, I cannot think of any way at all to find when one of these individual terms is equal to or less than 1/100.

My goal right now is finding the first term that will equal 1/100 basically... but I do not see any way this is possible. If there is a way I can find it please help me; if I should be going a completely different way to solve this problem please let me know. I've been thinking about this problem for a few hours and just cannot come up with anyway of finding the answer. Thanks for any help!

 

[Mark44]

Homework Statement

The problem says approximate the sum to within 1/100 of the value of the infinite sum, and the sum is
1/2-(1x3)/(2x4)+(1x3x5)/(2x4x6) and so on... (Teacher said I can leave answer in summation notation so I just need to find how many terms I need to add together.)

Homework Equations

n/a

The Attempt at a Solution

Since it is an alternating series, I would just need to find the first term that is equal to or less than 1/100 to know how many of the terms I need to add together and the terms are (1x3x...x(2n+1))/(2x4x...(2n+2)). However, I cannot think of any way at all to find when one of these individual terms is equal to or less than 1/100.

My goal right now is finding the first term that will equal 1/100 basically... but I do not see any way this is possible. If there is a way I can find it please help me; if I should be going a completely different way to solve this problem please let me know. I've been thinking about this problem for a few hours and just cannot come up with anyway of finding the answer. Thanks for any help!

Can you write the general term of the series? If so, use it to figure out what n needs to be so that the general term is < .01.

 

[Incand]

If you write $(2*4*6*\dots) = 2(1*2*3*\dots)$ Is there any terms that cancel each other?

 

[member 508213]

@Mark44 The General Term of the series would just be (-1)^n(1x3x...x(2n+1))/(2x4x...x(2n+2)!) but I can't really use this algebraically to find the value for n

 

[Mark44]

@Mark44 The General Term of the series would just be (-1)^n(1x3x...x(2n+1))/(2x4x...x(2n+2)!) but I can't really use this algebraically to find the value for n

No, that's not what it is. You shouldn't have that factorial in there. $2 \cdot 4 \cdot 6 \dots \cdot (2n + 2) = 2(n + 1)!$. Maybe that's what you were shooting for.

Anyway, write your inequality (can omit the $(-1)^n$) and solve. Typically the way to solve inequalities like this one is to try a few values of n to see when you get a true statement.

 

[member 508213]

@Mark44 I just accidently put in the ! I didn't mean to

 

[member 508213]

No, that's not what it is. You shouldn't have that factorial in there. $2 \cdot 4 \cdot 6 \dots \cdot (2n + 2) = 2(n + 1)!$. Maybe that's what you were shooting for.

Anyway, write your inequality (can omit the $(-1)^n$) and solve. Typically the way to solve inequalities like this one is to try a few values of n to see when you get a true statement.

Also... 2*4*6*8 doesn't equal 2(n+1)! it would equal 2^n(n+1)!

 

[Mark44]

Also... 2*4*6*8 doesn't equal 2(n+1)! it would equal 2^n(n+1)!

Yes, my mistake -- you are right, almost. 2⋅4⋅6⋯⋅(2n+2)=2^{n + 1}(n+1)!

 

[member 508213]

Yes, my mistake -- you are right, almost. 2⋅4⋅6⋯⋅(2n+2)=2^{n + 1}(n+1)!

but I can rewrite 2x4x6x...(2n+1) as 2^n(n!) but rewriting 1x3x5... seems a little more difficult

 

[member 508213]

but I can rewrite 2x4x6x...(2n+1) as 2^n(n!) but rewriting 1x3x5... seems a little more difficult

should probably be 2^(n+1)(n+1)! for 2x4x6... my mistake

 

[Ray Vickson]

Homework Statement

The problem says approximate the sum to within 1/100 of the value of the infinite sum, and the sum is
1/2-(1x3)/(2x4)+(1x3x5)/(2x4x6) and so on... (Teacher said I can leave answer in summation notation so I just need to find how many terms I need to add together.)

Homework Equations

n/a

The Attempt at a Solution

Since it is an alternating series, I would just need to find the first term that is equal to or less than 1/100 to know how many of the terms I need to add together and the terms are (1x3x...x(2n+1))/(2x4x...(2n+2)). However, I cannot think of any way at all to find when one of these individual terms is equal to or less than 1/100.

My goal right now is finding the first term that will equal 1/100 basically... but I do not see any way this is possible. If there is a way I can find it please help me; if I should be going a completely different way to solve this problem please let me know. I've been thinking about this problem for a few hours and just cannot come up with anyway of finding the answer. Thanks for any help!

Computationally, it might be easiest to not try to get a general formula and then figure out the value of $n$ from that. It might be more straightforward to write the series as
S = t_1 - t_2 + t_3 - \cdots
where
t_1 = \frac{1}{2}\; \; \text{and}\; \; t_n = \frac{2n-1}{2n} t_{n-1}, \; n \geq 2
You can just keep calculating the $t_n$ until $t_n < 0.01$ (for example, in a spreadsheet). Doing that is also useful in revealing just how slowly the series converges, and how many terms you need to keep to have any kind of decent accuracy. For example, 1000 terms are not nearly enough to give you $t_n < 0.01$.

However, if you really want a formula for $t_n$ you can apply the result
2 \times 4 \times 6 \times \cdots \times 2k = 2^k k!
to the formula
t_n = \frac{1 \times 3 \times \cdots \times 2n-1}{2 \times 4 \times \cdots \times 2n}<br /> = \frac{(2n-1)!}{(2 \times 4 \times \cdots \times (2n-2) )(2 \times 4 \times \cdots \times 2n)}
This will give you a formula involving factorials and powers of 2 only.