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Math 256 Quiz 6 Solutions Calculus II Kate Bella 1) Find a formula for the nth term of the sequence 1 2 3 4 0, , , , ..... . 4 6 8 10 We know that we need each term of the sequence to be calculated with a formula using its position as an input. Notice that the numerator is one less than the position of the term. Also, we see that the denominator is always an even number. We want this set of even numbers to start at 2, which means that it will always be double the position. So we have the formula an = n−1 . 2n 2) Find the limit of the sequence 2n + 1 . 3n We can consider the limit of the corresponding rational function, and apply l’Hopital’s rule. lim n→∞ 3) Show that the sequence 3 n2 2n + 1 2 2 = lim = n→∞ 3 3n 3 converges by showing that it is bounded and monotonic. First, we can see that this function is decreasing with the following inequalities. n < n+1 2 < (n + 1)2 1 (n + 1)2 3 (n + 1)2 n 1 n2 3 n2 > > Also, notice that all of the terms will be positive since the denominator is squared, so there is a lower boundary of 0. The upper bound will be 3 since this is the first value of the sequence, and we already know it is strictly decreasing. Thus the sequence is bounded.