# convergence review.pdf - Calc 2 Series...

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Calc 2 - Series Convergence/Divergence Review This sheet reviews some important ideas and tests from sections 10.2 to 10.6. Remember, lim n →∞ f ( n ) = L means “As n gets bigger and bigger, f ( n ) gets closer and closer to L .” (This is not a rigorous definition, but an intuitive one.) Also, lim n →∞ f ( n ) = is shorthand for “As n gets bigger and bigger, f ( n ) gets bigger and bigger.” 1 Limits The most basic limit is lim n →∞ 1 n = 0. Also, if 0 < A < 1 < B then lim n →∞ A n = 0 and lim n →∞ B n = . Theorem (L’Hopital’s Rule) . Suppose f ( x ) and g ( x ) are continuous functions. (i) If lim n →∞ f ( n ) = 0 and lim n →∞ g ( n ) = 0 , then lim n →∞ f ( n ) g ( n ) = lim n →∞ f 0 ( n ) g 0 ( n ) . (ii) If lim n →∞ f ( n ) = ±∞ and lim n →∞ g ( n ) = ±∞ , then lim n →∞ f ( n ) g ( n ) = lim n →∞ f 0 ( n ) g 0 ( n ) . Note: L’Hopital’s Rule only works if the limit is an indeterminate form that looks like 0 0 or ± . There are other indeterminate forms, and if they show up, you need to rearrange the limit somehow: 0 · ∞ ∞ - ∞ 0 0 1 0 Example 1 Evaluate lim n →∞ ln( n ) n . Solution: Since lim n →∞ ln( n ) = and lim n →∞ n = , then we can use L’Hopital’s Rule: lim n →∞ ln( n ) n = lim n →∞ 1 /n 1 = lim n →∞ 1 n = 0 . Theorem. If g ( x ) is a function and f ( x ) is a continuous function, then lim n →∞ f ( g ( n )) = f ( lim n →∞ g ( n )) . This is helpful for finding new limits. Example 2 Evaluate lim n →∞ n n . Solution: Recall n n = n 1 n . The limit lim n →∞ n 1 n is an indeterminate form that looks like 0 , so we need to rearrange the limit somehow. Since e x and ln( x ) are inverse functions, then f ( n ) = e ( ln( f ( n )) ) is true for any function f ( x ). Using this, we get lim n →∞ n 1 n = lim n →∞ e ln( n 1 n ) . Since ln( a m ) = m ln( a ), then we get lim n →∞ e ln( n 1 n ) = lim n →∞ e ln( n ) n . Since e x is a continuous function, then lim n →∞ e ln( n ) n = e lim n →∞ ln( n ) n = e 0 = 1. Putting this all together, we get lim n →∞ n n = 1. 1
Example 3 Let t 6 = 0 be some number. Evaluate lim n →∞ ( 1 + t n ) n . Solution: Note lim n →∞ ( 1 + t n ) n is an indeterminate form that looks like 1 , so we need to rearrange the limit somehow. Using a similar approach as the last example, lim n →∞ 1 + t n n = lim n →∞ e ln ( ( 1+ t n ) n ) = lim n →∞ e n ln ( 1+ t n ) = e ( lim n →∞ n ln ( 1+ t n ) ) Now we only need to evaluate lim n →∞ n ln ( 1 + t n ) . Since lim n →∞ 1 n = 0 and lim n →∞ ln(1 + t n ) = ln(1 + 0) = ln(1) = 0, then we can use L’Hopital’s Rule: lim n →∞ n ln(1 + t n ) = lim n →∞ ln(1 + t n ) 1 /n = lim n →∞ - t/n 2 1+ t n - 1 /n 2 = lim n →∞ t 1+ t n 1 = lim n →∞ t 1 + t n = t. Putting this all together, we get lim n →∞ ( 1 + t n ) n = e t . Theorem (Sandwich Theorem) . Let f ( x ) , g ( x ) and h ( x ) be functions. Suppose there is a large number N such that g ( n ) f ( n ) h ( n ) is true for all n N .