# How do you use the Ratio Test on the series #sum_(n=1)^oo(-10)^n/(4^(2n+1)(n+1))# ?

By Ratio Test, the posted series converges absolutely.

By Ratio Test:

By canceling out common factors:

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To use the Ratio Test on the series ( \sum_{n=1}^\infty \frac{(-10)^n}{4^{2n+1}(n+1)} ), we evaluate the limit:

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

where ( a_n ) represents the ( n )-th term of the series.

In this case, ( a_n = \frac{(-10)^n}{4^{2n+1}(n+1)} ).

Now, let's find ( a_{n+1} ):

[ a_{n+1} = \frac{(-10)^{n+1}}{4^{2(n+1)+1}((n+1)+1)} = \frac{(-10)^{n+1}}{4^{2n+3}(n+2)} ]

Now, let's find the ratio ( \frac{a_{n+1}}{a_n} ):

[ \frac{a_{n+1}}{a_n} = \frac{\frac{(-10)^{n+1}}{4^{2n+3}(n+2)}}{\frac{(-10)^n}{4^{2n+1}(n+1)}} ]

[ = \frac{(-10)^{n+1}}{4^{2n+3}(n+2)} \cdot \frac{4^{2n+1}(n+1)}{(-10)^n} ]

[ = \frac{(-10)(n+1)}{4^2(n+2)} ]

Now, let's take the limit as ( n ) approaches infinity:

[ \lim_{n \to \infty} \frac{(-10)(n+1)}{4^2(n+2)} = \lim_{n \to \infty} \frac{-10n - 10}{16n + 32} = \frac{-10}{16} = -\frac{5}{8} ]

Since the absolute value of this ratio is less than 1, specifically ( \frac{5}{8} ), by the Ratio Test, the given series ( \sum_{n=1}^\infty \frac{(-10)^n}{4^{2n+1}(n+1)} ) 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.

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