X Values Convergence Evaluation
- Thread starter NaN-Gram
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Steven G
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- Dec 30, 2014
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You have a few small mistakes. You really should look at your book for the formula.
1st mistake is that you should take the nth root of [the absolute value of what comes after the summation symbol]. This means that you do not have to deal with the alternating sign part, namely (-1)n or (-1)n+1.
Now you need to compute the limit of [the absolute value of what comes after the summation symbol] as x goes to infinity.
More formally given \(\displaystyle \Sigma_0^\infty \dfrac{(-1)^n(x+2)^n}{n}\)
To use the root test to test for convergence you need to compute \(\displaystyle \lim_{n\to\infty}|(\frac{(x+2)^n}{n})^{1/n}|\)
Now you are not done yet! If want this series to converge then you want this limit to less than 1. Since the limit will be in terms of x maybe you can restrict x to make the limit less than 1.
Please give this a try.
1st mistake is that you should take the nth root of [the absolute value of what comes after the summation symbol]. This means that you do not have to deal with the alternating sign part, namely (-1)n or (-1)n+1.
Now you need to compute the limit of [the absolute value of what comes after the summation symbol] as x goes to infinity.
More formally given \(\displaystyle \Sigma_0^\infty \dfrac{(-1)^n(x+2)^n}{n}\)
To use the root test to test for convergence you need to compute \(\displaystyle \lim_{n\to\infty}|(\frac{(x+2)^n}{n})^{1/n}|\)
Now you are not done yet! If want this series to converge then you want this limit to less than 1. Since the limit will be in terms of x maybe you can restrict x to make the limit less than 1.
Please give this a try.