- #1

Badmouton

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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

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- Thread starter Badmouton
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- #1

Badmouton

- 7

- 0

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

- #2

- 3,475

- 257

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

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

- #3

damabo

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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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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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Yes, the wolframAlpha link confirms that it's the same problem.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

@Badmouton,

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

This equivalent to (2*7

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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