Find Sum of Series 2/n*7^n for n=1 to ∞

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SUMMARY

The sum of the series 2/n*7^n for n=1 to ∞ can be expressed as S(x) = ∑(n=1 to ∞) (2/(n*7^n)). This series converges and can be evaluated using the formula for the Taylor series expansion of the natural logarithm, specifically S(x) = -2 * ln(1 - x) with x = 1/7. The final result of the series is confirmed through Wolfram Alpha as equivalent to the problem discussed in the referenced Physics Forums thread.

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you better use \frac{a}{b} to make clear what fraction you want to make.
make sure to use [/itex ] (but without the latter space before the bracket) to place the expression you want between those two.
 
Badmouton said:
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
S(x) = \sum_{n=1}^{\infty} \frac{x^n}{n} and then substituting the correct value of x.

RGV
 
jbunniii said:
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.

\displaystyle \sum_{n=1}^{\infty} \frac{2}{n7^{n}}

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

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

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