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For each of the following sequences determine whether they converge, and if they do, find their limit. Prove your answers. You are allowed to use the statements proven and examples done in class. (a) n+1 (−1) n n ​ (b) ( n+1 ​ − n ​ ) n ​ (c) ( n+1 ​ − n ​ )n. (d) (3 n ​ ) 1/2n . (e) a n +b n a n+1 +b n+1 ​ , where 0

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Answer and Step-by-step explanation:

It seems like you’re asking about the convergence of several sequences. Here are my answers:

(a) The sequence a_n = (n+1)(-1)^n/n is an alternating sequence. The limit of the absolute value of the sequence |a_n| = (n+1)/n is 1, which is not equal to 0. Therefore, by the Alternating Series Test, the sequence a_n does not converge.

(b) The sequence b_n = ((n+1)/n - n/n)^n can be simplified to b_n = (1 + 1/n)^n. This sequence converges to Euler’s number e ≈ 2.71828.

© The sequence c_n = ((n+1)/n - n/n)^n can be simplified to c_n = (1 + 1/n)^n. This sequence converges to Euler’s number e ≈ 2.71828.

(d) The sequence d_n = (3^n)^(1/(2n)) can be simplified to d_n = 3^(1/2). Since this expression does not depend on n, the sequence converges to 3^(1/2).

(e) For the sequence e_n = (a^n + b^n)/(a^(n+1) + b^(n+1)), where 0 < a < b, we can use L’Hopital’s Rule to find that the limit of the sequence is equal to the limit of (na^(n-1)b^n)/(a^nb^(n-1)(n+1)), which simplifies to (nb)/(a(n+1)). As n approaches infinity, this expression approaches 0. Therefore, the sequence converges to 0.

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