anonymous
  • anonymous
Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to an error < 0.001. Problem: (-1)^n/n5^n from n=1 to infinity.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Okay, first to show that this series is convergent, we can employ the Ratio test or Root test. I will do the ratio test which states the following: \[L=\lim_{n \rightarrow \infty}|a _{n+1}/a_{n}|\] If the limit of the series is less than 1, then the series is absolutely convergent In order to do this we perform the following limit: \[\lim_{n \rightarrow \infty}|[(-1)^{n+1}/(n+1)5^{n+1}]\times [n5^{n}/(-1)^{n}]|\]Simplify by canceling the (-1)^n and the 5^(n)\[\lim_{n \rightarrow \infty}|[(-1)^{n}(-1)/(n+1)5^{n}(5)]\times [n5^{n}/(-1)^{n}]|\]\[\lim_{n \rightarrow \infty}|[(-n)/5(n+1)|\]\[1/5\lim_{n \rightarrow \infty}n/(n+1)=1/5<1\]Using L'Hospitals rule for the limit of n/(n+1) So, according to the ratio test the limit is absolutely convergent. I'll answer the next part in a following response.
anonymous
  • anonymous
Do you happen to know the exact value of this series? Otherwise it would be extremely difficult to know how many terms you need to get such an error.
anonymous
  • anonymous
I know the series is decreasing, however in order to get an error of 0.001 the exact sum needs to be known. I tried solving for n, but I'm not sure how. If anyone could help answering the rest of this. I would appreciate it.

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anonymous
  • anonymous
Okay, I think I figured it out. Since \[\sum_{n=1}^{\infty}(-1)^{n}/(n5^{n})\]Is a alternating series, it can be shown by the Alternating Series Error Estimate theorem that \[\left| b_{n} \right| \le a_{n+1}\] Where \[b_{n} = 1/(n5^{n})\]\[a_{n+1}=0.001\]Meaning the non-alternating term summation has to be less than or equal to the error. So, if you keeping going through the terms of the sequence using the formula above until the statement is true, i.e.\[b_{3}=1/(3\times5^{3})=1/375 \ge 0.001\]\[b_{4}=1/(4\times5^{4})=1/2500=0.0004 \le 0.001\]then the number of terms needed to add to find the error less than 0.001 is 4 terms.

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