The sum of a series

  • Thread starter Badmouton
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  • #1
Badmouton
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  • #2
jbunniii
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  • #3
damabo
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you better use \frac{a}{b} to make clear what fraction you want to make.
make sure to use [itex][/itex ] (but without the latter space before the bracket) to place the expression you want between those two.
 
  • #4
Ray Vickson
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Serie's sum of 2/n*7^n?
How do we find it's sum for n=1 to n=inf?

I really do not know how to start, wolfram alpha gave me the answer, but I'm not making any sense out of it.
http://www.wolframalpha.com/input/?i=n=1+to+n=inf+2/(n*7^n)+sum

You do it by finding the sum
[tex] S(x) = \sum_{n=1}^{\infty} \frac{x^n}{n}[/tex] and then substituting the correct value of x.

RGV
 
  • #5
SammyS
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Is it the same as the problem in this thread? I'm not sure because you didn't use any parentheses in your expression.

https://www.physicsforums.com/showthread.php?t=637936
Yes, the wolframAlpha link confirms that it's the same problem.

@Badmouton,

The expression you are summing, with the proper set of parentheses: 2/(n*7n).

This equivalent to (2*7-n)/n also equivalent to (2*(1/7)n)/n .

All of these are more readable using LaTeX, which allows you to include the summation symbol, Ʃ , along with the summation limits.

[itex]\displaystyle \sum_{n=1}^{\infty} \frac{2}{n7^{n}}[/itex]

[itex]\displaystyle \sum_{n=1}^{\infty} \frac{2\left(7^{-n}\right)}{n}[/itex]

[itex]\displaystyle \sum_{n=1}^{\infty} \frac{2\left(\frac{1}{7}\right)^{n}}{n}[/itex]
 

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