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

  • Thread starter Badmouton
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In summary, the conversation discusses finding the sum of the series 2/n*7^n for n=1 to n=inf. The user is unsure how to approach the problem and references a Wolfram Alpha link for the answer. Another user confirms that it is the same problem as discussed in a previous thread and suggests using LaTeX for better readability.
  • #1
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][/itex ] (but without the latter space before the bracket) to place the expression you want between those two.
 
  • #4
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
[tex] S(x) = \sum_{n=1}^{\infty} \frac{x^n}{n}[/tex] and then substituting the correct value of x.

RGV
 
  • #5
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.

[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]
 

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

1. How do I find the sum of the series 2/n*7^n for n=1 to ∞?

To find the sum of this series, you can use the formula for the sum of an infinite geometric series, which is a/(1-r), where 'a' is the first term and 'r' is the common ratio. In this case, the first term is 2 and the common ratio is 7. Therefore, the sum of the series is 2/(1-7) = -1/3.

2. What is the pattern in this series?

This series follows a geometric pattern, where each term is multiplied by 7 and the denominator increases by 1. So, the first term is 2, the second term is 2*7 = 14, the third term is 14*7 = 98, and so on.

3. Is this series convergent or divergent?

This series is convergent, meaning that as n approaches infinity, the sum of the series approaches a finite number. In this case, the sum of the series is -1/3, which is a finite number.

4. How can I prove that this series is convergent?

To prove that this series is convergent, you can use the ratio test. The ratio test states that if the limit of |a(n+1)/a(n)| is less than 1, then the series is convergent. In this case, the limit is 7/2, which is less than 1, so the series is convergent.

5. Can this series be used in real-world applications?

Yes, this series can be used in various real-world applications, such as in finance to calculate compound interest or in physics to model exponential growth or decay. It can also be used in computer science to approximate certain algorithms and in statistics to model data trends.

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