Math  /  Calculus

QuestionMATH102 (Calculus II) First exam - Page 2 of 4 April 25, 2024
1. (21/2\left(21 / 2\right. points) The sum of the series k=05k3k2k\sum_{k=0}^{\infty} \frac{5^{k} 3^{-k}}{2^{k}} is A. 6 B. 5 C. -6 D. -5
2. ( 21/221 / 2 points) One of the following values of pp makes the series k=0pkk2k3+1\sum_{k=0}^{\infty} \frac{p^{k} k^{2}}{k^{3}+1} conditionally convergent. A. -1 B. 1 C. 2 D. 3
3. (2 1/21 / 2 points) The series k=0(1)k3k2+1\sum_{k=0}^{\infty}(-1)^{k} \frac{3}{k^{2}+1} is A. conditionally convergent. B. divergent. C. absolutely convergent.
4. ( 21/221 / 2 points) Which of the following series is convergent? A. k=1kk2+4\sum_{k=1}^{\infty} \frac{k}{k^{2}+4} B. k=11k+3\sum_{k=1}^{\infty} \frac{1}{k+3} kk2+4<nn2=1n×nn2+4>n2n1=12ndim1k+3<1k+3>12k Jir \begin{array}{l} \frac{k}{k^{2}+4}<\frac{n}{n^{2}}=\frac{1}{n} \times \frac{n}{n^{2}+4}>\frac{n}{2 n^{1}}=\frac{1}{2 n} \mathrm{dim} \\ \frac{1}{k+3}<\frac{1}{k+3}>\frac{1}{2 k} \text { Jir } \end{array} C. k=11k(1+ln2(k))\sum_{k=1}^{\infty} \frac{1}{k\left(1+\ln ^{2}(k)\right)} D. k=1(1)k4k+33k5\sum_{k=1}^{\infty}(-1)^{k} \frac{4 k+3}{3 k-5}
5. ( 21/221 / 2 points) One of the following values of pp makes the series n=1n!pn((n+1)!+1)\sum_{n=1}^{\infty} \frac{n!}{p^{n}((n+1)!+1)} convergent A. 2 B. 0.2 C. 0.5 D. 1 lim(n+1)!Pn+1((n+1+1)!+1)Pn((n+1)!+1)n!=(n+1)n!P2P((n+2)(n+1)!+1)PD2((n+1)!+1)=ln(n+1)((n+1)!+1P((n+2)(n+1)!\begin{aligned} \lim & \frac{(n+1)!}{P^{n+1}((n+1+1)!+1)} \cdot \frac{P^{n}((n+1)!+1)}{n!} \\ & =\frac{(n+1) n!}{P^{2} \cdot P((n+2)(n+1)!+1)} P^{D^{2}((n+1)!+1)}=\ln \frac{(n+1)((n+1)!+1}{P((n+2)(n+1)!} \end{aligned} lim(n+1)((n+1)!+1)P((n+2)!+1)=lim(n+1)((n+1)!+1)(n+1)!+(nP(n+2)!+P1Plim(n+1)((n+1)!+1)(n+2)!+1)P((n+2)!+1)P(n+2)!+P\begin{array}{l} \lim \frac{(n+1) \cdot((n+1)!+1)}{P((n+2)!+1)}=\lim \frac{(n+1)((n+1)!+1)(n+1)!+(n}{P(n+2)!+P} \\ \left|\frac{1}{P}\right| \lim \frac{(n+1)((n+1)!+1)}{(n+2)!+1)} \quad P((n+2)!+1) \quad P(n+2)!+P \end{array}

Studdy Solution
Using the ratio test, we find that p=2p = 2 makes the series converge.
The solutions to the problems are:
1. A. 6
2. A. -1
3. A. conditionally convergent.
4. C. k=11k(1+ln2(k))\sum_{k=1}^{\infty} \frac{1}{k(1+\ln^2(k))}
5. A. 2

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