lim as n --> +infinity sigma (top n) (bottom k=1) (3/4)^k Solution We can split

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We can split up this limit to give: lim (n-->infinity) sum(i=1 to n) [(i/3)^3 + 1]/n => lim (n-->infinity) sum(i=1 to n) [(i/3)^3/n + 1/n] = lim (n-->infinity) [1/n * sum(i=1 to n) (i/3)^3 + 1/n * sum(i=1 to n) 1]. But since sum(i=1 to n) 1 = n, we have: lim (n-->infinity) [1/n * sum(i=1 to n) (i/3)^3 + 1/n * sum(i=1 to n) 1] = lim (n-->infinity) [1/n * sum(i=1 to n) (i/3)^3 + (1/n)(n)] = 1 + lim (n-->infinity) [1/n * sum(i=1 to n) (i/3)^3] = 1 + 1/27 * lim (n-->infinity) [1/n * sum(i=1 to n) i^3]. Using the summation formula sum(i=1 to n) i^3 = n^2(n + 1)^2/4, we have: 1 + 1/27 * lim (n-->infinity) [1/n * sum(i=1 to n) i^3] = 1 + 1/27 * lim (n-->infinity) [1/n * n^2(n + 1)^2/4] = 1 + 1/108 * lim (n-->infinity) n(n + 1)^2 = +infinity.

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