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Numerade Educator



Problem 43 Hard Difficulty

For which of the following series is the Ratio Test inconclusive (that is, if fails to give a definite answer)?

(a) $ \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n^3} $

(b) $ \displaystyle \sum_{n = 1}^{\infty} \frac {n}{n^2} $

(c) $ \displaystyle \sum_{n = 1}^{\infty} \frac {( - 3)^{n-1}}{\sqrt{n}} $

(d) $ \displaystyle \sum_{n = 1}^{\infty} \frac {\sqrt{n}}{1 + n^2} $


a. The Ratio Test is inconclusive.
b. The Ratio Test is conclusive.
c. The Ratio Test is conclusive.
d. inconclusive


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Video Transcript

three to the Siri's. Let's determine whether the ratio test is inconclusive. Suffer part. Ay, let's look at this one first. This will be our end So we'd like to look at the limit as an goes to infinity. Absolute value a n plus one over n So in our case, here's an plus one and then we'LL divide that by an Now let's go ahead and multiply out This will give us and cubed over and plus one Cute Now, if you like, you could go ahead and use Lopez House rule. In any case, you'LL get one when you evaluate this limit and we know that the ratio test is inconclusive when the limit equals one Since our limit equals one in part a we conclude that the ratio test is inconclusive for the first series And now let's go on to party Now this is our man and we could even go ahead and do a little cancellation to right, this is one over it. So we'LL have the limit n goes to infinity once again and plus one over a n Now the numerator This is one over and plus one and the denominator one over it going to flip that denominator and then multiply it once again? You could use Lopez House rule here and you'LL get a limit of one just like in party and we conclude that the ratio test is inconclusive. Here is well, so that's for a party. Now let's go on to Parsi Here. This is our A So once again we start off with the usual limit and absolute value A and plus one over, eh? So we'LL have n plus one minus one over radical and plus one and divide that by a m So now notice that we can ignore the absolute value If we remove the negative signs we could drop the absolute value So what I'LL do here is just take this on to the next page so I could have some more room. So we're still in party, so have three to the end over radical and plus one and then radical end over three the and minus one. So go ahead and cancel a cz Many threes as you can you could cross off and minus one You'LL still have one three up top and then this limit over here we'LL go to one. So this limit becomes three, which is bigger than one. So Siri's see diverges, and in particular, we see that the ratio test is not inconclusive. Here. That's for party. So it's the first one that we come across to, where the ratio test was not inconclusive and let's see what part visas. So this Siri's was from one to infinity radical, one plus and square in the bottom. So now we look at the limit a n plus one over n. So it's not writing this out. So here's my a n plus one, and divide that by a M. So, as usual, we'LL go ahead and take that second fraction, flip it upside down and multiply it, and we can verify that both of these limits are one. So for the first fraction, if it helps you Khun, rewrite this as and plus one over n one plus one over N, And when you take the limit that goes toe one plus zero and the radical, which is just radical one equals one, and similarly, for the second fraction, you can divide by and square or use low Patel's rule. But this limit will also go to one so that we conclude that the limit goes toe one. And so that the ratios has his inconclusive for party. Hi.