Tính biểu đạt và hiệu suất của ngôn ngữ?

I, = ∫01∫0sin-1rr2er4cos t dtdr = ∫01[r2er4sin t]0sin-1r dr. = ∫01r3er4 dr = [er4/4]01 = (e - 1)/4. Write r1 = m1/n1 and r2 = m2/n2, where m1, n1, m2, n2 ... Oct 9, 2020 ... Improving previous answers, DiscreteLimit[Normal[Series[Cos[Pi*Sqrt[4 n^2 + 5 n + 1]],{n,Infinity,1}]],n -> Infinity] (*-(1/Sqrt[2])*). (n+1)(n+2). ; 0 k n;. 0. ; k>n for all k; n 2 N. A short calculation gives the ... is, f lim(Ax)n = l. Since C = (cnk) is a strongly regular triangle ... ... 1 ∞ ( − 1 ) n − 1 1 n 2 converges absolutely.. Definition 6.55 ... \(\lim\limits_{n\to\infty} \dfrac{3n+4}{2n^2+3n+5} = 0\text{.}\) If we let \(f ... Example 2: To determine whether the series ∞∑n=14n2n+3n converges or diverges, we'll look for a series that "behaves like" it when n is large. Solution 2: Since ... HINT: If we take logarithm of this expression we obtain n1​(ln(1+2/n)+ln(1+4/n)+…ln(1+(2n)/n)). converges (p-series with p = 2 > 1). So the given series converges too, by the Comparison Test. Or when using the Limit Comparison Test. Example. Test the ... (a) (10 points). n e−n sin(1 n. ) o. ∞ n=1. Solution: (b) (10 points) n. 1. 2+(-1)n o. ∞ n=1. Solution: PLEAE TURN OVER. Page 3. Math 1B. Practice First Midterm ... Jul 12, 2015 ... We call such an a the limit of the sequence and write limn→∞ an = a. Proposition 1. The sequence 1 n converges to zero. Proof. Let > 0. We ... tίve integer n, an = 1 —(n + l)~p then. 1. lim al = 0 , n-*oo. 2. Σ α? exists and oo. 3. Σ (1 —anY exists . n—Q. Proof. To establish 1, lim al = lim (1 - n~p)n.