MATH 1502 Midterm: MATH 1502 GT C2T201sol
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I: a) comparison test and the bound k2. 1 k2 yields convergence: this is an alternating sum. 1 ln(k) we see that ln(2k) (ln(k))2 tends to zero and is decreasing. Therefore the series converges: using the root test, we to compute lim n (cid:18) n. Ii: a) the series is a geometric series and converges precisely when which leads to. 7 < x < 3 for the interval of convergence: here we use the ratio test and get (k + 1)!|x|k+1 (2k + 2)! (2k)! |x|kk! (k + 1)|x| (2k + 2)(2k + 1) which converges to zero as k no matter how large x is. Thus, the interval of convergence is the whole real line: the key observation is that. K + 1 + k and hence using again the ratio test. 1 we see that this converges to |x| as k . Thus we know that the series converges for.
