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For each of the following series; find the sum or determine that it diverges(L1)" 5" I[ Select ] 20 rilL)z"Sellect |i...

Question

For each of the following series; find the sum or determine that it diverges(L1)" 5" I[ Select ] 20 rilL)z"Sellect |i

For each of the following series; find the sum or determine that it diverges (L1)" 5" I[ Select ] 20 ril L)z" Sellect |i



Answers

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.
$$\sum_{n=1}^{\infty} \frac{n}{n+2}$$

Okay. This question gives you a series 5 -5 Fords plus 5/4 squared minus 5/4 cubed. And if I take the ratio of each term to the term before it, I can tell that I'm multiplying by negative 1/4 every time. So the common ratio yeah equals negative 1/4. The absolute value of that common ratio Is less than one. So this infinite geometric series will converge converges to a 1/1 minus r. A one is the first term. Five Divided by 1 -3 common ratio, negative 1/4. His five divided by 5/4 which is four. The sum to infinity is positive for. And that's it for this problem. This problem comes up, comes out kind of quick. If you're not sure about the common ratio, remember that I can always do the second term divided by the first and then the third term divided by the second and so on. Oops hold on, I still get negative 1/4. And that needs to be a pattern that you can tell continues. Or in general I could take a sub N plus one, divided by a sub N. Mhm. If that number is a constant for the problem, then that's your common ratio. As long as the absolute value of your common ratio is less than one. Then I can use this formula, but only if your ratio is less than one. Be careful about that.

Indication would be calling about the geometric stress. Yes, which, under farm? The submission that I attempt the armor and from the infinity equal to the first time over one minutes are even Only if absolutely are, must be strictly smaller than one. Otherwise, it will be divergent. Okay, in this question, were given the five minus 5/4 5/4 square, minus five off for about three. So on here, notice that we can turn it into the submission. Yeah, we should get in the five here terms when the man is one of the far starting from zero to infinity. Well, and yeah, And now, clearly, we can identify. This will be the geometric series with the ICO to five. There are equal absolute r equals one of a far smaller than one. Therefore, we can apply this formula to get equal with you the first time we see equal to the five here, dividing by the one plus one of a far. Then we get equal to the 20/5, and then we get echo to the far. So I just want me to answer

Your Grace should test where I need your computer leading it off Investing ability. I am. This one will be told him six time after Newt. You in terms the to embarrass Joon the very by the five and rest Jew in the street. Tonja They went in, but they end will be multiplied by the reciprocal final and a scorn and rescue if Antonio invited by longtime six. After that, you and and now we can cancel over time here with one of this one and ah fei about. And this one with concern with this power here. And we have answered the embassy off. Attorney counselor, with this material and we have left without a limit. And Justin impunity, You know, we have the number five year complete Know somebody one of five. And inside we have the I can't and Lis Judy binding by an industry. And it's Angus to infinity. Tell limit in San Diego studio job number one. They could get you over five smaller than one. Therefore, this seriously, it would be a convergent

In discussion, we have our cities one upon to plus two upon 3 plus three upon 4 plus four upon 5 plus and so on, plus and upon And plus one plus so on. We need to find whether the series converges or diverges and we need to find it to some if the series converges. So that's how this whole discussion we know that. Mhm. If the sequence oh partial some has a limit as and tends to infinity, then we can conclude that D cities converges otherwise diverges from the given cities. We can observe that one upon 2 is less than to a boundary. two upon 3 is less than three upon four, three Upon 4 is less than four upon 5. So we can conclude that the ratio is increasing But still it is always there's 10 1 hands. We can see that the term is approaching two one and Since the term is not approaching 20, therefore, we can conclude that the cities one upon to plus two up on three plus three upon four, plus so on, plus And upon and plus one and so on diverges. So this is a final answer for this problem. I hope you understand revolution. Thank you


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