Series CheatSheet

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Proposition 2.2. Let f(x) be a real function on the interval [1, +∞) which is positive, continuous and decreasing. Let n be a natural number, the following inequality holds: n+1 1 f(x) dx ≤ n f(i) ≤ i=1 n+1 1 f(x) dx + f(1) − f(n) The series +∞ i=1 f(i) is convergent if and only if the improper integral +∞ 1 f(x) dx is convergent. 3rd method: With the following proposition we can compare a given series with some other series that we have already studied: Proposition 2.3. Let an, bn be two positive sequences. Suppose that there is a natural number n such that for all n > n the following inequality holds: an ≤ bn Under these assumptions if the series +∞ n=1 bn is convergent, the series +∞ n=1 an is also convergent. If the series +∞ n=1 an is divergent, the series +∞ n=1 bn is also divergent. 4th method: The following technique is the basic tool to turn a complicated series into something simpler. It’s not the ultimate technique (there are cases where it’s inconclusive) but it’s the sharpest tool that we have (for the moment). Proposition 2.4. Let an, bn be positive sequences. Suppose that the sequence bn converges to a real number C > 0. The series +∞ n=1 anbn is convergent if and only if the series +∞ n=1 an is convergent. If the limit of bn is zero we have a weaker statement: Proposition 2.5. Let an, bn be positive sequences. Suppose that the sequence bn converges to zero. If the series +∞ n=1 an is convergent, the series +∞ n=1 anbn is also convergent. 3 Fundamental limits The following limits of sequences are fundamental. These should be used together with the limit comparison technique. • • • • lim n→+∞ lim n→+∞ lim n→+∞ sin 1 n 1 n 1 − cos 1 n 1 n 2 = 1 = 1 2 ln(1 + 1 n ) 1 = 1 n e lim n→+∞ 1 n − 1 1 = 1 n 2
4 Strategy There is no algorithm to determine the convergence of a series, nonetheless we can try to outline a strategy that would work in many circumstances. Step 1: If the sequence an doesn’t converge to zero the series is divergent and we are done. Step 2: If the limit is zero the series can be convergent divergent or might not have a limit. If possible, we apply limit comparison repeatedly to replace transcendental functions with rational functions. We replace in the following way: • • • • sin 1 1 ≈ n n cos 1 1 ≈ 1 − n 2n2 ln(1 + 1 1 ) ≈ n n e 1 n ≈ 1 + 1 n If after replacing the transcendental pieces the sequence is equal to zero, limit comparison is inconclusive!!! Step 3: If the sequence is an algebraic function (polynomials, divisions and radicals only) we collect the highest power of n and we apply limit comparison again. Step 4: If everything worked fine the series should be a fundamental series (one in the list below) or more in general a product/quotient of the following functions: n p , a n , ln q (n), n!, n n . At this point we can try to conclude the calculation using the integral test, the 2 k -test or comparison. Example 4.1. Determine for which positive real numbers α, β the following series is convergent: +∞ n=1 1 sin nβ ln α (n 2 + 1) We apply limit comparison and replace sin and we obtain the following equivalent series: +∞ n=1 We extract the leading term from the logarithm: +∞ n=1 1 n β lnα (n 2 + 1) 1 nβ 2 ln(n) + ln 1 + 1 n2 α We don’t need to approximate ln 1 + 1 n2 since we are summing it to ln(n) which is divergent. We apply limit comparison again: +∞ (ln(n)) α n=2 3 n β
Page 1: 1 Definitions and properties Series
Page 5 and 6: • • • It is convergent. Done

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