Determine if the following series converges, diverges, or is inconclusive. ∑ n = 1 ∞ ( 2 n 2 − 1 n 2 + 5 ) − n \sum_{n=1}^{\infty}\left(\frac{2n^2-1}{n^2+5}\right)^{-n}

this problem, we're asked to determine if the following series converges, diverges, or is inconclusive. And we have the series that goes from n equals one to infinity of two n squared minus one over n squared plus five, all raised to the negative n. Now, right at first, this almost looks like a case where we could use the root test, since I see we have this raised to a power here with n in it, but notice that we have negative n rather than n. Well, it turns out even in this case, we still actually can use the root test, because this would be the same thing as having this whole sum from n equals one to infinity of, in this case, we could just go ahead and flip this because this would be the same thing as just taking one over two n squared minus one over n squared plus five. And notice that this would be raised to a power of n. And so if this was also raised to a power of n, then we could just see that this whole thing is raised to a power of n. And And we can totally do that because, in this case, one raised to any power would just still be one, so we can raise this whole thing to the n power. And this right here would be our a sub n to the n. So, yes, we can use the root test even in cases like this. So So let's go ahead and try this. Now our root test, we're going to see that we have the limit as n approaches infinity for the nth root of the absolute value of a sub n raised to a power of n. That's what we have right here for the root test, and this whole thing is gonna equal r. Now, there are some rules when it comes to the root test. If r is less than one, we're going to have our case where our series converges. If r is greater than one, then this means the series is going to diverge. And if r is equal to one, then we could say our series is inconclusive. So let's see what happens when we evaluate this limit. So to do this limit here, it's going to be the limit as n approaches infinity, and what we have up here is one over two n squared minus one over n squared plus five. Just to make things easier on myself, I'm just going to take this fraction and flip it and bring it to the top. So what it's going to give me is the nth root of the absolute value of my a sub n, which I can see flipping this a sub n to the n power, that's going to be n squared plus five over two n squared minus one. And then I can brace this all to a power of n. Now what I can do is take a look and see how this is going to behave as we go from one to infinity. And as we go to infinity for this n right here, well, notice that this is going to be positive. Because even if we had a one in there, we would just get two times one squared minus one, which would just give us positive one. In the long run, as we go to infinity, this is always going to be positive. This is just gonna keep getting bigger and bigger. So because of that, we can go ahead and get rid of these absolute value bars. And we can also recognize that now the root here is going to cancel. The root n is gonna cancel with the n power there, leaving us with the limit as n approaches infinity of n squared plus five over two n squared minus one. Now in this case, what I can do is determine what's going to happen as n goes to infinity. And notice for this case, what I want to do is I want to look at where the highest exponent on n is. You can see we have an n squared on top and an n squared on bottom. Since we have two as the highest exponent in the numerator and denominator, I'll go ahead and do a ratio of the coefficients in front of the n squareds. So that's just going to give me one half. Now notice that r is equal to one half, we went ahead and did this limit. And what I can do is recognize that any case where r is less than one means our series is going to converge. One half is clearly less than one, so our series converges, and that is the solution to this problem. Hope you found this helpful.

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

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Multiple Choice

Determine if the following series converges, diverges, or is inconclusive.
$\sum_{n = 1}^{\infty} {(\frac{2 n^{2} - 1}{n^{2} + 5})}^{- n}$

Converges

Diverges

Inconclusive

Verified step by step guidance

Step 1: Analyze the given series ∑_{n=1}^{∞} $\left$($\frac{2n^2-1}{n^2+5}\[\right$)^{-n}. The first step is to simplify the general term $\left$($\frac{2n^2-1}{n^2+5}\]\right$)^{-n}. Rewrite the term as $\left$($\frac{n^2(2 - \frac{1}{n^2}$)}{n^2(1 + $\frac{5}{n^2}$)}$\right$)^{-n}, which simplifies to $\left$($\frac{2 - \frac{1}{n^2}$}{1 + $\frac{5}{n^2}$}$\right$)^{-n}.

Step 2: Examine the behavior of the base $\frac{2 - \frac{1}{n^2}$}{1 + $\frac{5}{n^2}$} as n approaches infinity. As n becomes very large, $\frac{1}{n^2}$ and $\frac{5}{n^2}$ approach 0, so the base approaches $\frac{2}{1}$ = 2.

Step 3: Consider the exponent -n. Since the base approaches 2 and the exponent is -n, the term $\left$($\frac{2 - \frac{1}{n^2}$}{1 + $\frac{5}{n^2}$}$\right$)^{-n} behaves like 2^{-n} for large n.

Step 4: Recall that the series ∑_{n=1}^{∞} 2^{-n} is a geometric series with a common ratio r = $\frac{1}{2}$, which satisfies |r| < 1. Therefore, the geometric series converges.

Step 5: Conclude that the given series ∑_{n=1}^{∞} $\left$($\frac{2n^2-1}{n^2+5}$$\right$)^{-n} converges because its terms asymptotically behave like a convergent geometric series.

5:44m

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Struggling with Calculus?

Determine if the following series converges, diverges, or is inconclusive. ∑n=1∞(2n2−1n2+5)−n\(\sum\)_{n=1}^{\(\infty\)}\(\left\)(\(\frac{2n^2-1}{n^2+5}\]\right\))^{-n}

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Determine if the following series converges, diverges, or is inconclusive.
$\sum_{n = 1}^{\infty} {(\frac{2 n^{2} - 1}{n^{2} + 5})}^{- n}$