I'm trying to prep for a test this following week and want to make sure if I'm doing these correctly.
1) ∞sigma n= 1 (n^(3/2)+5)/(3^n +2).
For this series, I used the direct comparison test. I used 1/n^1.5. Since 1/n^1.5 converges, so does the original series.
2) ∞sigma n= 0 (2^n + 5)/(3^(n) +2). For this one, I used direct comparison again. I used (2/3)^n. This also converges so the original series will converge since (2^n + 5)/(3^(n) +2) < (2/3)^n
3) ∞sigma n= 0 (2^n (n+1)^n)/(n^n). I used the ratio test and got (2(n+1))/n. n+1/n clearly diverges since the numerator is greater than the denominator.
4) ∞sigma n= 1 (-1^n) * (n^3 + 3)/( n^3+1). This is an alternating series. (n+1)^3 +3 / (n+1)^3 +1 < original series. So it clearly decreases. But the limit as n goes to infinity is not 0, so it diverges? I'm not too sure about this one.
1) ∞sigma n= 1 (n^(3/2)+5)/(3^n +2).
For this series, I used the direct comparison test. I used 1/n^1.5. Since 1/n^1.5 converges, so does the original series.
2) ∞sigma n= 0 (2^n + 5)/(3^(n) +2). For this one, I used direct comparison again. I used (2/3)^n. This also converges so the original series will converge since (2^n + 5)/(3^(n) +2) < (2/3)^n
3) ∞sigma n= 0 (2^n (n+1)^n)/(n^n). I used the ratio test and got (2(n+1))/n. n+1/n clearly diverges since the numerator is greater than the denominator.
4) ∞sigma n= 1 (-1^n) * (n^3 + 3)/( n^3+1). This is an alternating series. (n+1)^3 +3 / (n+1)^3 +1 < original series. So it clearly decreases. But the limit as n goes to infinity is not 0, so it diverges? I'm not too sure about this one.
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1. comparison with n^(3/2)/(3^n). use the ratio test on these to find out if this converges or diverges. your answer will be converges by the LIMIT comparison test.
2. you can't really say that (2/3)^n is larger than the original series. say limit comparison test.
3. wrong. This diverges because the limit as n approaches infinty is 2 which is greater than 1.
4. this diverges by the n-th term test because it bounces between 1 and -1.
2. you can't really say that (2/3)^n is larger than the original series. say limit comparison test.
3. wrong. This diverges because the limit as n approaches infinty is 2 which is greater than 1.
4. this diverges by the n-th term test because it bounces between 1 and -1.
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Wait, idk why I said 4 diverges by the alternating series test, it diverges for the test of divergence or nth term test.
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Bang on.
5 does not converge: as n goes to ∞, the terms alternate between -1 and 1 .... and you should know what happens when that happens :-). Hint: the answer depends on how the terms are picked .....
5 does not converge: as n goes to ∞, the terms alternate between -1 and 1 .... and you should know what happens when that happens :-). Hint: the answer depends on how the terms are picked .....