Determine the convergence or divergence of the series. ∑ n = 1 ∞ 5 n − 1 2 n \sum_{n=1}^{\infty}\frac{5^{n-1}}{2^{n}}

this problem, we are asked to determine the convergence or divergence of the series that we have. Now notice we're not told what type of series this is or what tests we need to use, so all of this we're going to have to figure out. So I went ahead and copied this flowchart down here that we've been using to see how we can go about determining what series we're going to need and how to determine the convergence. So here we have this sum from n equals one to infinity of five to the n minus one over two n. So how can I go about solving this problem? Well, I'm gonna go ahead and start with the divergence test. The divergence test is a pretty straightforward one, but it doesn't always come up as something that's gonna be divergent. It's either it's gonna be a divergent or it's gonna be inconclusive, in which case we need to move on to another test. So in this case, I'm gonna take the limit as n approaches infinity. And what I need to do is take this limit of a sub n. Now what is my a sub n? Well, in this case, it's gonna be everything that we have inside the sum right here. Now sometimes you can just tell how n is going to behave as it approaches infinity. But in this case, it actually looks a little more complicated since we have this five to the n minus one over two to the n power. So I'm actually going to write all of this inside here. So we're gonna have five to n minus one over two to the power of n. This is our whole ace of n right here. Now what I need to do from here is figure out how this is going to behave. And the way that I can do that is by going ahead and rewriting what we have. So we have the limit as n approaches infinity. This is not gonna change until we apply the limit. And then five to the n minus one, this is the same thing as five to the n times five to the negative one, which I can go ahead and just rewrite this as that being over five. Then It's gonna be multiplied by one over two n since we still have this two n in the denominator. So what this is going to give me is, in this case, we're going to have one or this limit still. We need to keep writing on this limit. You don't wanna forget this. So we have the limit, and that's going to be one over five, because we have the one over five right there, times five to the n over two to the two to the n. And we can actually rewrite this whole thing as just five over two to the power of n since both are raised to a power of n. Now at this point, it may become a little bit more clear what's going to happen as n approaches infinity. Notice here that it's just this exponent going to infinity, which is just gonna make this number bigger and bigger. Notice since we have a heavier numerator right here, it is gonna be a larger and larger number as we go to infinity. So because of that, this whole thing is just going to blow up to infinity. And I noticed we got a case where the limit did not come out to zero. We actually did get this result of infinity that came out. So because of that, that means the divergence test actually did work, and this series does indeed diverge. So diverge would be the solution to this problem. Hope you found this helpful, and let's move on.

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14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

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Multiple Choice

Determine the convergence or divergence of the series.
$\sum_{n = 1}^{\infty} \frac{5^{n - 1}}{2^{n}}$

Inconclusive

Diverges

Converges

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Verified step by step guidance

Step 1: Recognize that the given series is a geometric series. A geometric series has the form $ \sum_{n=1}^{\infty} ar^{n-1} $, where $ a $ is the first term and $ r $ is the common ratio.

Step 2: Rewrite the given series $ \sum_{n=1}^{\infty} \frac{5^{n-1}}{2^{n}} $ to match the geometric series form. Factor out $ \frac{1}{2} $ from the denominator to express the series as $ \sum_{n=1}^{\infty} \left( \frac{5}{2} \right)^{n-1} \cdot \frac{1}{2} $.

Step 3: Identify the first term $ a $ and the common ratio $ r $. Here, $ a = \frac{1}{2} $ and $ r = \frac{5}{2} $.

Step 4: Recall the convergence criterion for a geometric series: A geometric series converges if and only if $ |r| < 1 $. Check whether $ |r| = \left| \frac{5}{2} \right| $ satisfies this condition.

Step 5: Conclude whether the series converges or diverges based on the value of $ |r| $. If $ |r| \geq 1 $, the series diverges; otherwise, it converges.