Infinite series, does ∑ n/2^n diverge?

  • #1

Homework Statement



[tex]\sum^{\infty}_{n=1} \frac{n}{2^{n}}[/tex]

Does this series converge or diverge?

Homework Equations





The Attempt at a Solution



By the Cauchy condensation test (http://en.wikipedia.org/wiki/Cauchy_condensation_test) I think this one diverges. But not sure if I am using it correctly.

According to the test,

[tex]\sum^{\infty}_{n=1} \frac{n}{2^{n}}[/tex]

converges if and only if

[tex]\sum^{\infty}_{n=1} 2^{n} \frac{2^{n}}{2^{n}} = \sum^{\infty}_{n=1} 2^{n}[/tex]

converges, which doesn't.

Thank you for any help.
 

Answers and Replies

  • #2
34,897
6,639

Homework Statement



[tex]\sum^{\infty}_{n=1} \frac{n}{2^{n}}[/tex]

Does this series converge or diverge?

Homework Equations





The Attempt at a Solution



By the Cauchy condensation test (http://en.wikipedia.org/wiki/Cauchy_condensation_test) I think this one diverges. But not sure if I am using it correctly.

According to the test,

[tex]\sum^{\infty}_{n=1} \frac{n}{2^{n}}[/tex]

converges if and only if

[tex]\sum^{\infty}_{n=1} 2^{n} \frac{2^{n}}{2^{n}} = \sum^{\infty}_{n=1} 2^{n}[/tex]

converges, which doesn't.
In the line above, you aren't using the condensation test correctly. For your series, f(n) = n/(2n), so what would be f(2n)?

A test that would be simpler to apply would be the Limit Ratio Test.
Thank you for any help.
 
  • #3
In the line above, you aren't using the condensation test correctly. For your series, f(n) = n/(2n), so what would be f(2n)?

A test that would be simpler to apply would be the Limit Ratio Test.

Ah, okay, I see it where I went wrong..

Using the limit ratio test

[tex] lim \left| a_{n+1}/a_{n} \right| = lim \left| \frac{(n+1)/2^{n+1}}{n/2^{n}} \right| = 1/2 < 1 [/tex]

So it converges...
 
  • #4
34,897
6,639
Yes, indeed.
 

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