Determine whether the series converges or diverges: Solution 1) 8 ? (4 + 3^n) /

Question

Determine whether the series converges or diverges:

Explanation / Answer

1) 8 ? (4 + 3^n) / 2^n n=1 use the limit comparison test as the following: check two rules: an = (4 + 3^n) / 2^n & bn = (3^n/2^n) = (3/2)^n 1) lim bn n--> 8 lim (3/2)^n = 8 divergent and it divergent by geometric series since 3/2 > 1 n--> 8 at this point, we are going to stop here since the b_n diverges. therefore, it diverges by the limit comparison test. ============== 2) 8 ? (9^n) / (3 + 10^n) n=1 use the limit comparison test as the following: check two rules: an = (9^n) / (3 + 10^n) & bn = (9^n/10^n) = (9/10)^n a) lim bn n--> 8 lim (9/10)^n = 0 convergent, and it converges by geometric series since 9/10 < 1 n--> 8 b) lim an / bn n--> 8 lim [ (9^n) / (3 + 10^n) ] / [ (9^n / 10^n) ] n--> 8 lim [ (9^n) / (3 + 10^n) ] * [ 10^n/9^n ] n--> 8 lim [ 10^n / (3 + 10^n) ] ====> large in charge n--> 8 lim [ 10^n / 10^n ] n--> 8 lim 1 = 1 > 0 ( convergent ) n--> 8 since the ? bn converges and ? an/bn converges too, then the ? an from n = 1 to 8 is convergent by the limit comparison test. =============== 3) 8 ? (1 + 4^n) / (1 + 3^n) n=1 use the limit comparison test as the following: check two rules: an = (1 + 4^n) / (1 + 3^n) & bn = (4^n/3^n) = (4/3)^n 1) lim bn n--> 8 lim (4/3)^n = 8 divergent and it divergent by geometric series since 4/3 > 1 n--> 8 at this point, we are going to stop here since the b_n diverges. therefore, it diverges by the limit comparison test.

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