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Series Convergence/Divergence Questions

  • #1
Determine if the following converge or diverge and explain why.

1. [tex]\Sigma_{n=1}^{\infty} \frac{\sin{\frac{n\pi}{2}}}{n}[/tex]

2. [tex]\Sigma_{n=1}^{\infty} \frac{\ln{n}}{n^2}[/tex]


For (1) I know that the numerator is bounded by -1 and 1 but do not know how to solve it from there.

For (2) I used the comparison test and did the following:

[tex]\frac{\ln{n}}{n^2} < \frac{n}{n^2} = \frac{1}{n}[/tex]

and then

[tex]\Sigma_{n=1}^{\infty} \frac{1}{n}[/tex] diverges (harmonic series) therefore (2) diverges by Comparison Test.

Is this the proper way to solve (2)?
 

Answers and Replies

  • #2
Dick
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For the first one, write out the first 5 or 6 terms of the series numerically and see if that gives you any ideas. For the second one, no, bounding a series above by a divergent series doesn't prove it diverges. 1/n^2<1/n also, but 1/n^2 converges. Can you think of an upper bound comparison that converges?
 
  • #3
Okay I got (2), (thanks for the correction), by comparing (2) to [tex]\frac{\sqrt{n}}{n^2}=\frac{1}{n^\frac{3}{2}}[/tex] which converges (p-series w/ p>1), therefore (2) converges. Is this correct?

As for (1) I wrote out the terms and discovered it is an alternating series with the even terms all equal to 0 as follows:

1-1/3+1/5-1/7+...

I can't seem to write a series to satisfy the requirements however...any guidance would be appreciated.
 
  • #4
Dick
Science Advisor
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Okay I got (2), (thanks for the correction), by comparing (2) to [tex]\frac{\sqrt{n}}{n^2}=\frac{1}{n^\frac{3}{2}}[/tex] which converges (p-series w/ p>1), therefore (2) converges. Is this correct?

As for (1) I wrote out the terms and discovered it is an alternating series with the even terms all equal to 0 as follows:

1-1/3+1/5-1/7+...

I can't seem to write a series to satisfy the requirements however...any guidance would be appreciated.
Yes, that's right for (2). For (1) you have an alternating series with decreasing terms. How about the alternating series test?
 
  • #5
Yea I figured it had to do with that.

If we define the alternating series to be: [tex]\Sigma_{n=1}^{\infty} (-1)^{n-1}b_n[/tex]

then I can't seem to determine a [tex]b_n[/tex] which defines the series for all n, especially since every odd [tex]n[/tex] changes sign and all the even [tex]n[/tex]'s are 0...I must be missing something...
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
Yea I figured it had to do with that.

If we define the alternating series to be: [tex]\Sigma_{n=1}^{\infty} (-1)^{n-1}b_n[/tex]

then I can't seem to determine a [tex]b_n[/tex] which defines the series for all n, especially since every odd [tex]n[/tex] changes sign and all the even [tex]n[/tex]'s are 0...I must be missing something...
Your original series is not strictly an alternating series, true. But just throw out the zero terms. That makes your series equal to (-1)^(n-1)/(2n-1), right?
 
  • #7
Ah, the something I was missing was the fact that we were not explicitly using the values in the original series, which was confusing me. Instead we just defined a completely new series whose sum is equivalent to the old one. Thank you so much for finally making that clear...I should have known but was just caught up in finding something that worked! Many thanks!
 

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