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

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Homework Help Overview

The discussion revolves around finding the sum of the series \( \frac{2}{n \cdot 7^n} \) for \( n=1 \) to \( \infty \). Participants express uncertainty about how to approach the problem and reference external tools for assistance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants question how to start solving the series and whether it is similar to another problem discussed in a different thread. There are suggestions to clarify the expression using proper notation and LaTeX formatting. One participant mentions a potential approach involving a known series sum.

Discussion Status

The discussion is ongoing, with participants seeking clarification and exploring different interpretations of the series expression. Some guidance on notation has been provided, but no consensus or resolution has been reached.

Contextual Notes

There is mention of confusion regarding the expression due to the lack of parentheses, which may affect understanding. Participants also reference external resources for verification of the problem's similarity to another thread.

Badmouton
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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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