Finding the sum of finite terms of a Maclaurin series

ohpoonet
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Homework Statement



Hi. Find the sum of the first 10 terms in the series:

1, (ln2)/1!, (ln2)^2/2!, (ln^3)/3!...

Homework Equations



I guess the Maclaurin series, which is e^k = 1, k/1!, (k^2)/2!, (k^3)/3!...
In my case, k = ln2.

The Attempt at a Solution



I tried to use the sum of the first n terms Sn = (u1(r^n -1))/(r-1). However I cannot find r because it is not constant.

I realize if you replace k with ln2, you get e^(ln2) = 2 = 1 + k/1! + (k^2)/2!...So the series approaches 2.

As a last resort I will go caveman style and add up all terms from 0-10. That's a lot of decimals, so if anyone can help, it will be appreciated.
 
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Here's what Maple gives for the writing on the caveman's wall: 0.99999999952832752678

Interestingly close to 1. It also gives a formula for the sum of n terms in terms of an incomplete gamma function with parameter ln(2), which somehow I doubt is what you are looking for.
 
Haha thanks BUT,

Do you agree that the series approaches 2? after all, e^ln2 is 2. If so how can the first 10 terms add up to 1?

I never heard of Maple until now, it seems like an algebra application. My question is now, did you solve for S of 10? in this case, what did you use for r, the ratio? Or did it simply add up all the terms 'manually'?

Sorry but i don't know about Maple. I will research the gamma function though I do not think its the right direction.

More help will be great. Thanks in advance!
 
ohpoonet said:
Haha thanks BUT,

Do you agree that the series approaches 2? after all, e^ln2 is 2. If so how can the first 10 terms add up to 1?

I never heard of Maple until now, it seems like an algebra application. My question is now, did you solve for S of 10? in this case, what did you use for r, the ratio? Or did it simply add up all the terms 'manually'?

Sorry but i don't know about Maple. I will research the gamma function though I do not C
More help will be great. Thanks in advance!

e^(ln(2)) may be 2, but your series doesn't start with the first term of the expansion of e^x. The '1' is missing. LCKurtz just added them up using Maple and didn't solve for anything.
 
ohpoonet said:
Haha thanks BUT,


I never heard of Maple until now, it seems like an algebra application. My question is now, did you solve for S of 10? in this case, what did you use for r, the ratio? Or did it simply add up all the terms 'manually'?

Sorry but i don't know about Maple.

Maple is a sophisticated and powerful mathematics program with many uses. It has no problem adding up lots of terms keeping many decimal places; that's all I did. As Dick has pointed out, the first term of 1 in the exponential expansion is missing from your series which is why it approximates 1. The fact that it is so close after 10 terms is due to the high rate of convergence caused by the factorial in the denominator.

Many schools have a license agreement for a student version of Maple. You can read about it at:

http://www.maplesoft.com/
 
Hi guys,

Thanks for the response. I understand now. I actually used MS excel to add up the terms. I also found out that in:

((xLNa)^n)/n!

the graph approaches a^x.

(For example i put x=2 and a=3 and found the sum of the first 9 terms approached 9.)

Just for thought.

Thanks again.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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