# How do you test for convergence of #Sigma n e^-n# from #n=[1,oo)#?

See below.

and

because

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The series:

is convergent.

We can determine the convergence of the series:

using the ratio test:

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The series is convergent by the ratio test.

Since the ratio test gives a value less than one, the series is convergent by the ratio test.

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To test for the convergence of the series ( \sum_{n=1}^{\infty} ne^{-n} ), you can use the ratio test.

[ a_n = ne^{-n} ]

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

[ \lim_{n \to \infty} \left| \frac{(n+1)e^{-(n+1)}}{ne^{-n}} \right| ]

[ \lim_{n \to \infty} \left| \frac{n+1}{n} \cdot \frac{e^{-n}}{e^{-(n+1)}} \right| ]

[ \lim_{n \to \infty} \left| \frac{n+1}{n} \cdot \frac{e}{1} \right| ]

[ \lim_{n \to \infty} \left| \frac{n+1}{n} \cdot e \right| ]

[ \lim_{n \to \infty} e = e ]

Since the limit is a constant value ( e ) which is less than 1, by the ratio test, the series ( \sum_{n=1}^{\infty} ne^{-n} ) converges.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- How do you find the radius of convergence of the power series #Sigma 2^n n^3 x^n# from #n=[0,oo)#?
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- How do you know if the summation #(-12)^n/n# where n is between 3 to infinity is convergent or divergent?
- How do you use the integral test to determine if #Sigma lnn/n^3# from #[2,oo)# is convergent or divergent?

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