# How do you use the comparison test for #sum 1/(5n^2+5)# for n=1 to #n=oo#?

The Direct Comparison Test (not to be confused with the Limit Comparison Test) says, paraphrased:

Notice how the larger series determines convergence, not the smaller one. It's exactly like this:

If b, then a. If not a, then not b.

which is a statement and then its contrapositive, which must be true. The affirmative is the convergence, and the denial is the divergence.

This is not true:

If b, then a. If a, then b.

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To use the comparison test for the series (\sum \frac{1}{5n^2 + 5}) from (n = 1) to (n = \infty), we need to find a series whose terms are less than or equal to the terms of the given series and whose sum is known.

We can rewrite the given series as (\sum \frac{1}{5(n^2 + 1)}). Now, since (n^2 + 1 > n^2) for all (n \geq 1), we have (\frac{1}{5(n^2 + 1)} < \frac{1}{5n^2}) for all (n \geq 1).

The series (\sum \frac{1}{5n^2}) is a well-known convergent series, known as a p-series with (p = 2). Its sum is (\frac{1}{5} \times \frac{\pi^2}{6}).

By the comparison test, since (\frac{1}{5(n^2 + 1)} < \frac{1}{5n^2}) for all (n \geq 1) and (\sum \frac{1}{5n^2}) converges, then (\sum \frac{1}{5(n^2 + 1)}) also 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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