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[tex]\sum_{n=0}^{\infty}ne^{(-nx)}[/tex]

I found this problem on the internet and couldn't solve it. Thanks for your help.

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[tex]\sum_{n=0}^{\infty}ne^{(-nx)}[/tex]

I found this problem on the internet and couldn't solve it. Thanks for your help.

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How about the integral test for convergence? For what values of x would the required integral converge?alexmcavoy@gmail.com said:

[tex]\sum_{n=0}^{\infty}ne^{(-nx)}[/tex]

I found this problem on the internet and couldn't solve it. Thanks for your help.

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That is what I am having trouble with. I actually got this problem off of the MIT OpenCourseWare website under "Analysis I" and am looking to see how someone would solve this.

When I evaluate [tex]\sum_{n=0}^{\infty}ne^{(-n)}[/tex] on my calculator, I get [tex]\frac{1}{4(\sinh{(\frac{1}{2})})^{2}}[/tex]. How was that calculated?

When I evaluate [tex]\sum_{n=0}^{\infty}ne^{(-n)}[/tex] on my calculator, I get [tex]\frac{1}{4(\sinh{(\frac{1}{2})})^{2}}[/tex]. How was that calculated?

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It's related to a geometric series. Can you findalexmcavoy@gmail.com said:When I evaluate [tex]\sum_{n=0}^{\infty}ne^{(-n)}[/tex] on my calculator, I get [tex]\frac{1}{4(\sinh{(\frac{1}{2})})^{2}}[/tex]. How was that calculated?

[tex]\sum_{n=0}^{\infty}nx^n[/tex]

(where it converges of course-actually if you can find where this guy converges then you should be able to convert back to your original problem without difficulty)

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I guess I can't. How would you do this? I know the [tex]\frac{1}{1-x}[/tex] answer for the value of an infinite series, but don't know how to apply it to this particular problem.

I see that if I did find that value, let's say [tex]x=a[/tex], then I could replace [tex]x[/tex] with [tex]e^{(-x)}[/tex] and solve getting [tex]x=-\ln{(a)}[/tex].

Would this be correct?

I see that if I did find that value, let's say [tex]x=a[/tex], then I could replace [tex]x[/tex] with [tex]e^{(-x)}[/tex] and solve getting [tex]x=-\ln{(a)}[/tex].

Would this be correct?

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[tex]\lim_{n\rightarrow\infty}\frac{|(n+1)x^{(n+1)}|}{|nx^{(n)}|}=x[/tex] so [tex]-1<x<1[/tex]. So now I replace [tex]x[/tex] with [tex]e^{(-x)}[/tex]. This is where I am stuck because I can't solve it. What did I do incorrectly?

Thanks.

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well to find out for what values this series converges for is not particularly hard its just a straight application of the ratio test the same we would apply it to the power series to find the radius of convergence

[tex]\lim_{n\rightarrow\infty}\frac{(n+1)e^{(nx)}}{ ne^{(n+1)x}}=\frac{1}{e^x}<1[/tex]

[tex]1<e^{x}[/tex]

[tex]\log 1<\log e^{x}[/tex]

and so therefore the series converges for all [tex]x>0[/tex]

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Alex, you can do this.alexmcavoy@gmail.com said:I guess I can't. How would you do this?

Let:

[tex]w=\frac{1}{e^a}[/tex]

and:

[tex]z=\sum_{n=0}^{\infty}w^n[/tex]

Now, you know what that sum is right? It's an expression in terms of w.

Now take the derivative of both expressions with respect to w, adjust the sum to look like what you want and you're done.

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