Determine whether the given series is convergent. ∑ n = 1 ∞ 5 3 n − 7 \sum_{n=1}^{\infty}\frac{5}{3n-7}

Here we have part b of this problem that asks us to determine whether the given series is convergent, and what I can see here is that we have the sum from n equals one to infinity of five over three n minus seven. Now what I'm going to do when it comes to solving this is I'm first going to take this five that I see on top here, I'm going to pull it outside the summation. So that's going to give me five times the sum from n equals one to infinity of one over three n minus seven. Now why did I go ahead and do that? Well, this is actually a rule when it comes to constants. You can pull them outside of the sum, but notice the form that we have here. This very much follows the form of a general harmonic series. Because we know for a general harmonic series, it's gonna be in the form of one over a n plus b. And for a general harmonic series, this is always going to diverge. Now looking at this, I can clearly see that we do have this general form right here because our a is clearly three and then our b is this negative seven. So because of that, we can say that this is a general harmonic series and our series therefore must diverge. So divergent would be the solution to this problem, and that is how we can figure out whether this series is convergent or divergent. In this case, it's divergent. So let's now move on to case c.

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14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

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Multiple Choice

Determine whether the given series is convergent.
$\sum_{n = 1}^{\infty} \frac{5}{3 n - 7}$

It converges because it is a geometric series.

It converges because it is a general harmonic series.

It diverges because it is a general harmonic series.

It diverges because it is a geometric series.

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Verified step by step guidance

Step 1: Begin by analyzing the given series $ \sum_{n=1}^{\infty} \frac{5}{3n-7} $. Observe the general term $ a_n = \frac{5}{3n-7} $. This is neither a geometric series nor a harmonic series in its standard form.

Step 2: Recall that a geometric series has the form $ \sum_{n=0}^{\infty} ar^n $, where $ r $ is the common ratio. Check if the given series matches this form. It does not, so it is not a geometric series.

Step 3: Recall that a harmonic series has the form $ \sum_{n=1}^{\infty} \frac{1}{n} $, which diverges. The given series $ \frac{5}{3n-7} $ resembles a harmonic series but with a linear transformation in the denominator. Such transformations do not change the divergence behavior of the harmonic series.

Step 4: Apply the divergence test. If $ \lim_{n \to \infty} a_n \neq 0 $, the series diverges. Compute $ \lim_{n \to \infty} \frac{5}{3n-7} $. As $ n \to \infty $, the denominator grows indefinitely, so $ \lim_{n \to \infty} \frac{5}{3n-7} = 0 $. This does not confirm convergence; further analysis is needed.

Step 5: Conclude that the series diverges because it behaves similarly to a general harmonic series. The denominator $ 3n-7 $ grows linearly with $ n $, and such series are known to diverge.