# How do you use the direct comparison test to determine if #Sigma 3^n/(4^n+5)# from #[0,oo)# is convergent or divergent?

See explanation.

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To determine if the series ( \sum_{n=0}^{\infty} \frac{3^n}{4^n+5} ) converges or diverges, we can use the Direct Comparison Test.

First, observe that for ( n \geq 0 ):

[ \frac{3^n}{4^n + 5} < \frac{3^n}{4^n} = \left(\frac{3}{4}\right)^n ]

Since the series ( \sum_{n=0}^{\infty} \left(\frac{3}{4}\right)^n ) is a geometric series with ( |r| = \frac{3}{4} < 1 ), it converges.

Now, by the Direct Comparison Test, since ( 0 \leq \frac{3^n}{4^n + 5} < \left(\frac{3}{4}\right)^n ) and ( \sum_{n=0}^{\infty} \left(\frac{3}{4}\right)^n ) converges, then ( \sum_{n=0}^{\infty} \frac{3^n}{4^n + 5} ) also converges.

Therefore, the series ( \sum_{n=0}^{\infty} \frac{3^n}{4^n + 5} ) is convergent.

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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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