Series #9 - Q&A

Question 1 Find a formula for the nth partial sum of each series and use it to find the series' sum if the series converges. Answer 1: Given data: The series is: 9 10 + 9 100 + 9 1000 + …   . 10 9 ​ + 100 9 ​ + 1000 9 ​ +…. Consider the infinite series and compare it with the given series: 𝑏 1 + 𝑏 2 + 𝑏 3 + ⋯ + 𝑏 𝑛 . b 1 ​ +b 2 ​ +b 3 ​ +⋯+b n ​ . The nth partial sum of the given series is: 𝑠 𝑛 = ∑ 𝑘 = 1 𝑛 𝑏 𝑘 . s n ​ = k=1 ∑ n ​ b k ​ . The first term of the given series is: 𝑏 = 9 10 b= 10 9 ​ . The common ratio is: 𝑟 = 1 10 r= 10 1 ​ . The expression for the sum of series is: 𝑠 𝑛 = 𝑏 ( 1 − 𝑟 𝑛 ) 1 − 𝑟 . s n ​ = 1−r b(1−r n ) ​ . Substitute the values in the above expression: 𝑠 𝑛 = 9 10 ( 1 − ( 1 10 ) 𝑛 ) 1 − 1 10 = 9 10 ( 1 − ( 1 10 ) 𝑛 ) 9 10 . s n ​ = 1− 10 1 ​ 10 9 ​ (1−( 10 1 ​ ) n ) ​ = 10 9 ​ 10 9 ​ (1−( 10 1 ​ ) n ) ​ . Thus, the formula for the nth partial sum of series is: 𝑠 𝑛 = 100 99 ( 1 − 9 ( 1 100 ) 𝑛 ) . s n ​ = 99 100 ​ (1−9( 100 1 ​ ) n ). The expression for the limit of the nth partial sum of the given series is: 𝐿 = lim ⁡ 𝑛 → ∞ 𝑠 𝑛 = 100 99 ( 1 − 9 ( 1 100 ) ∞ ) . L= n→∞ lim ​ s n ​ = 99 100 ​ (1−9( 100 1 ​ ) ∞ ). The nth partial sum of the given series is: 1 11 . 11 1 ​ . The sequence of the partial sum converges to limit 𝐿 = 1 11 L= 11 1 ​ , so the given series converges. Thus, the sum of the series is: 9 10 + 9 100 + ⋯ = 1 11 . 10 9 ​ + 100 9 ​ +⋯= 11 1 ​ . Question 2 a. Another Maclaurin series we examined is for sin ⁡ ( 𝑥 ) sin(x). Find the series for sin ⁡ ( 𝑥 2 ) 2 2 sin(x 2 ) ​ . Express the result in summation notation, not just as a partial listing of the series terms. b. Another Maclaurin series we examined is for sin ⁡ ( 𝑥 ) sin(x): c. Find the series for sin ⁡ ( 𝑥 2 ) 𝑥 x sin(x 2 ) ​ . Answer 1: The Maclaurin series is the Taylor series expansion about point 𝑎 = 0 a=0. Find the value for 𝑓 ( 𝑥 ) f(x) at 𝑎 = 0 a=0: 𝑓 ( 𝑥 ) = sin ⁡ ( 𝑥 ) , 𝑓 ( 𝑎 ) = sin ⁡ ( 𝑎 ) , 𝑓 ( 0 ) = sin ⁡ ( 0 ) = 0. f(x)=sin(x),f(a)=sin(a),f(0)=sin(0)=0. Consider the formula for the Taylor series: 𝑓 ( 𝑥 ) = ∑ 𝑛 = 0 ∞ 𝑓 ( 𝑛 ) ( 𝑎 ) 𝑛 ! ( 𝑥 − 𝑎 ) 𝑛 . f(x)= n=0 ∑ ∞ ​ n! f (n) (a) ​ (x−a) n . Finding the values for the derivatives required to find the Taylor series: 𝑓 ( 𝑥 ) = sin ⁡ ( 𝑥 )    ⟹    𝑓 ′ ( 𝑥 ) = cos ⁡ ( 𝑥 ) , 𝑓 ′ ′ ( 𝑥 ) = − sin ⁡ ( 𝑥 ) , 𝑓 ′ ′ ′ ( 𝑥 ) = − cos ⁡ ( 𝑥 ) . f(x)=sin(x)⟹f ′ (x)=cos(x),f ′′ (x)=−sin(x),f ′′′ (x)=−cos(x). The Taylor series centered at 𝑎 = 0 a=0: sin ⁡ ( 𝑥 ) = 𝑥 − 𝑥 3 3 ! + 𝑥 5 5 ! − …   . sin(x)=x− 3! x 3 ​ + 5! x 5 ​ −…. For sin ⁡ ( 𝑥 2 ) 2 2 sin(x 2 ) ​ : sin ⁡ ( 𝑥 2 ) = 𝑥 2 − 𝑥 6 3 ! + 𝑥 10 5 ! − … , sin(x 2 )=x 2 − 3! x 6 ​ + 5! x 10 ​ −…, sin ⁡ ( 𝑥 2 ) 2 = 𝑥 2 2 − 𝑥 6 6 ! + 𝑥 10 2 ⋅ 5 ! − … , 2 sin(x 2 ) ​ = 2 x 2 ​ − 6! x 6 ​ + 2⋅5! x 10 ​ −…, = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 𝑥 2 + 4 𝑛 ( 2 𝑛 + 1 ) ! . = n=0 ∑ ∞ ​ (2n+1)! (−1) n x 2+4n ​ . Question 3 Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series. ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑛 2 . n=1 ∑ ∞ ​ n 2 (−1) n ​ . Answer 1: The given series is: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑛 2 . n=1 ∑ ∞ ​ n 2 (−1) n ​ . To check its convergence or divergence using the Alternating Series Test: Solution: The alternating series says that if we have series of form: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 − 1 𝑏 𝑛 , n=1 ∑ ∞ ​ (−1) n−1 b n ​ , then if, lim ⁡ 𝑛 → ∞ 𝑏 𝑛 = 0 lim n→∞ ​ b n ​ =0, and 𝑏 𝑛 b n ​ is a decreasing sequence, then the series ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 − 1 𝑏 𝑛 ∑ n=1 ∞ ​ (−1) n−1 b n ​ is said to be convergent. Since we have series: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑛 2 , n=1 ∑ ∞ ​ n 2 (−1) n ​ , we have sequence 𝑏 𝑛 = 1 𝑛 2 b n ​ = n 2 1 ​ . Now check both conditions using alternating series test for convergence: lim ⁡ 𝑛 → ∞ 𝑏 𝑛 = lim ⁡ 𝑛 → ∞ 1 𝑛 2 = 0 lim n→∞ ​ b n ​ =lim n→∞ ​ n 2 1 ​ =0, 𝑏 𝑛 > 𝑏 𝑛 + 1 b n ​ >b n+1 ​ . So, 𝑏 𝑛 b n ​ is a decreasing sequence as well. Since both conditions are satisfied, the given series is convergent. Question 4 Use the alternating series test to determine the convergence of the series ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 sin ⁡ 2 ( 𝑛 ) . n=1 ∑ ∞ ​ (−1) n sin 2 (n). Answer 1: We have given a series: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 sin ⁡ 2 ( 𝑛 ) . n=1 ∑ ∞ ​ (−1) n sin 2 (n). We know that for the alternating series test, series should be given in forms: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑎 𝑛 or ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 + 1 𝑎 𝑛 . n=1 ∑ ∞ ​ (−1) n a n ​ or n=1 ∑ ∞ ​ (−1) n+1 a n ​ . Our series is given in the form: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑎 𝑛 . n=1 ∑ ∞ ​ (−1) n a n ​ . Conditions for alternate series are: 𝑎 𝑛 > 0 a n ​ >0, and lim ⁡ 𝑛 → ∞ 𝑎 𝑛 = 0 lim n→∞ ​ a n ​ =0, and the series should be in decreasing order to be convergent. However, in this case: lim ⁡ 𝑛 → ∞ sin ⁡ 2 ( 𝑛 ) ≠ 0. n→∞ lim ​ sin 2 (n)  =0. Hence, the given series is not convergent. Question 5 Use the formula for the sum of a geometric series to find the sum or state that the series diverges (enter DIV for a divergent series): ∑ 𝑛 = 2 ∞ 5 𝑛 1 2 𝑛 . n=2 ∑ ∞ ​ 12 n 5 n ​ . Answer 1: Here, the given series is: ∑ 𝑛 = 2 ∞ 5 𝑛 1 2 𝑛 . n=2 ∑ ∞ ​ 12 n 5 n ​ . Let us expand the given series: ∑ 𝑛 = 2 ∞ ( 5 12 ) 𝑛 = ( 5 12 ) 2 + ( 5 12 ) 3 + ( 5 12 ) 4 + …   . n=2 ∑ ∞ ​ ( 12 5 ​ ) n =( 12 5 ​ ) 2 +( 12 5 ​ ) 3 +( 12 5 ​ ) 4 +…. We clearly see that the given series is an infinite geometric series with first term ( 5 12 ) 2 ( 12 5 ​ ) 2 and common ratio 5 12 12 5 ​ . Again, we know that the sum of an infinite geometric series with first term 𝑎 a (say) and common ratio 𝑟 ( ∣ 𝑟 ∣ ≤ 1 ) r(∣r∣≤1) is: 𝑆 = 𝑎 1 − 𝑟 . S= 1−r a ​ . So, 𝑆 = ( 5 12 ) 2 1 − 5 12 = 25 144 7 12 = 25 ⋅ 12 144 ⋅ 7 = 25 84 . S= 1− 12 5 ​ ( 12 5 ​ ) 2 ​ = 12 7 ​ 144 25 ​ ​ = 144⋅7 25⋅12 ​ = 84 25 ​ . Which is the required answer.