# Does #sum_1^oo [3/n(n+4)] - [(2^(2n)/7^n+1)]# converge?

Diverges.

Rearranging, the infinite seris is

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To determine if the series (\sum_{n=1}^{\infty} \left[\frac{3}{n(n+4)} - \frac{2^{2n}}{7^{n} + 1}\right]) converges or diverges, we need to analyze the convergence of its individual terms and apply convergence tests.

First, let's consider the term (\frac{3}{n(n+4)}). This term converges since it's a rational function with the degree of the numerator less than or equal to the degree of the denominator.

Next, let's examine the term (\frac{2^{2n}}{7^{n} + 1}). To determine its convergence, let's rewrite it as (\frac{(2^2)^n}{7^{n} + 1}) which simplifies to (\frac{4^n}{7^{n} + 1}). Now, we can use the ratio test.

Applying the ratio test:

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

[ = \lim_{n \to \infty} \left| \frac{\frac{4^{n+1}}{7^{n+1} + 1}}{\frac{4^n}{7^{n} + 1}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{4^{n+1}(7^n + 1)}{4^n(7^{n+1} + 1)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{4(7^n + 1)}{7(7^n + 1)} \right| ]

[ = \lim_{n \to \infty} \frac{4}{7} ]

[ = \frac{4}{7} ]

Since ( \frac{4}{7} < 1 ), by the ratio test, the series ( \sum_{n=1}^{\infty} \frac{2^{2n}}{7^{n} + 1} ) converges.

Therefore, since both terms converge, the series (\sum_{n=1}^{\infty} \left[\frac{3}{n(n+4)} - \frac{2^{2n}}{7^{n} + 1}\right]) 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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