anonymous
  • anonymous
Sublimation n=1 to infinity of (cos(n•pi))/(srt(n)). How do I solve this using the alternating series test? Why is a sub n greater than or equal to zero. The answer is converges, but I need help walking thru the steps. Thanks.
Mathematics
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anonymous
  • anonymous
Sublimation n=1 to infinity of (cos(n•pi))/(srt(n)). How do I solve this using the alternating series test? Why is a sub n greater than or equal to zero. The answer is converges, but I need help walking thru the steps. Thanks.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
You have a series,\[\sum_{n=1}^{\infty}\frac{\cos n \pi}{\sqrt{n}}\]It's important to realise that cos(n pi) will oscillate between -1 and 1 for each successive n. So your series is actually equal to\[\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\]
anonymous
  • anonymous
The alternating series test says, basically, for a series of the form,\[\sum_{n}(-1)^na_n\]the series will converge if \[\lim_{n \rightarrow \infty}a_n =0\]and \[a_{n+1}
anonymous
  • anonymous
Here, your a_n is\[a_n=\frac{1}{\sqrt{n}}\]which certainly goes to zero as n approaches infinity, and since sqrt(n) is monotonic increasing, it follows 1/sqrt(n) is monotonic decreasing, which is what you want.

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anonymous
  • anonymous
So, by the alternating series test, your series converges. Let me know if you need more help.

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