WebJan 24, 2024 · Step-by-step explanation: The given series is To find the sum of the series, we need to substitute the values for k in the series. Now, simplifying the square terms, we get, Multiplying the terms, Subtracting the values within the bracket term, we get, Now, adding all the terms, we get the sum of the series, Thus, the sum of the series is WebEvaluate the sum, ∑_ (k=1)^n 〖1/ (k (k+1) (k+2)……… (k+r)) 〗 742 views Jul 3, 2024 Evaluate the sum, ∑_ (k=1)^n 〖1/ (k (k+1) (k+2)……… (k+r)) 〗 Full Playlist Sequence and...
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WebFeb 6, 2024 · a. Find an upper bound for the remainder in terms of n. b. Find how many terms are needed to ensure that the remainder is less than 10 − 3 . c. Find lower and upper bounds (ln and Un, respectively) on the exact value of the series. ∑ k = 1 ∞ 1 3 k Answer & Explanation Neelam Wainwright Skilled 2024-02-07 Added 102 answers (a). WebDetermine whether the following series converges or diverges. In the case of convergence, state whether the convergence is conditional or ∑ k = 1 ∞ k 2 + 9 (− 1) k Choose the … rdbuf - in_avail
ρ — Adic Analogues of Ramanujan Type Formulas for 1/π
WebAbout this unit. This unit explores geometric series, which involve multiplying by a common ratio, as well as arithmetic series, which add a common difference each time. We'll get to … Web∑∞. n=k an diverges. If ρ = 1 , then we need a better test. Root test: Let an > 0 and ρ = limn→∞(an)1/n. If ρ < 1, then; ∑∞. n=k an (absolutely) converges. If ρ > 1, then. ∑∞. n=k an diverges. If ρ = 1 , then we need a better test. Alternating series test: If an is positive and decreases to 0 , then; ∑∞. n=k(− 1 ) nan ... Webk= X∞ n=k+1 1 n3 to be less than 0.05 =1 20 Now, the remainder estimate for the integral test says that Z∞ k+1 1 x3 dx ≤ R k≤ Z∞ k 1 x3 dx, so we know that R k<1 20whenever the integral on the right is less than20 In other words, we want 1 20 > Z∞ k 1 x3 dx = − 1 2x2 = 1 2k2 Now,1 2k2<20whenever 20 < 2k 2or, equivalently, when 10 < k2. sinbad the sailor cartoon 60\\u0027s