Midterm I for Calculu II (Solutions)

1. (1) $a_n=\frac{2^n+4^n}{3^n+4^n}=\frac{{\left(\frac{1}{2}\right)}^n+1}{{\left(\frac{3}{4}\right)}^n+1}\to 1$ as $n\to \mathrm{\infty }$

$\therefore a_n+0\therefore \sum a_n$ div

(2) $f(x)=\frac{1}{x(\ln x{)}^2},f(x)>0\mathrm{\&}\downarrow$ for $x⩾2$

$a_n=f(n)$ for $n\text{⧸}2$

${\int }_2^{\mathrm{\infty }}f(x)dx={\int }_2^{\mathrm{\infty }}\frac{1}{x(\ln x{)}^2}dx\underset{du=\frac{1}{x}dx}{\overset{=\frac{\ln x}{\mathrm{\infty }}{\int }_{\ln 2}^{\mathrm{\infty }}\frac{1}{u^2}du\mathrm{con}v.(p=2>1),(an>0)}{=}}$

$\therefore \sum a_n$ conv. & A.C. $(a_n>0)$

(3) $a_n=\frac{1}{n\sqrt[n]{n}},b_n=\frac{1}{n}$

$\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt[n]{n}}=1\Rightarrow$ both conv. or both div

$\because \sum b_n$ div. $\therefore \mathrm{\Sigma }{a}_n$ div

(4) $\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!(n+2)!(n+3)!}{(3n+3)!}\frac{(3n)!}{n!(n+1+1n+2)!}=\frac{(n+3)(n+2)(n+1)}{(3n+3)(3n+2)(3n+1)}\to \frac{1}{27}<1$

$\Rightarrow \sum a_n$ A.C

as $n\to \mathrm{\infty }$

(5) $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{U}_n,u_n=\sqrt{n^2+1}-n>0$,

$U_n=\frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{\sqrt{n^2+1}+n}=\frac{1}{\sqrt{n^2+1}+n}⩾$ in $n$

$\underset{n\to \mathrm{\infty }}{lim}{U}_n=0$. $\therefore$ it is conv. by alternating series tect

$\sum _{n=1}^{\mathrm{\infty }}{u}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{n^2+1}+n},b_n=\frac{1}{n}$

$\frac{u_n}{b_n}=\frac{n}{\sqrt{n^2+1}+n}\to \frac{1}{2}\therefore \sum u_n\mathrm{\&}\sum b_n$ buth conv. or both div. Since $\sum \frac{1}{n}$ div., $\sum U_n$ div. In summary, $\sum (-1{)}^n{U}_n$ is C.C

2. $\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x{|}^{n+1}}{\sqrt{(n+1{)}^2+3}}\cdot \frac{\sqrt{n^2+3}}{|x{|}^n}\to |x|$

$\therefore$ radius of cone. $r=1$

$x=1$ : $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{1}{\sqrt{n^2+3}}$ conv by alternating series test

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{n^2+3}}$ div. $\sin ce\frac{\frac{1}{\sqrt{n^2+3}}}{\frac{1}{n}}\to 1\sum \frac{1}{n}$ div

$\therefore$ C.C. at $x=1$

$x=-1:\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{n^2+3}}$ div

In summary, radius of conve is 1; interval of unv. is $(-1,1]$;

the series is A.C. in $(-1,1)$ & C.C. at $x=1$

3

$$\begin{array}{rl}f(x)& =(x+1)e^x=(x+1)[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots ]\\ & =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+\cdots \\ & =1+2x+(\frac{1}{2!}+1)x^2+(\frac{1}{3!}+\frac{1}{2!})x^3+\cdots \\ & =1+\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{(n-1)!}+\frac{1}{n!})x^n=\sum _{n=0}^{\mathrm{\infty }}\frac{n+1}{n!}{x}^n\end{array}$$

4. $\frac{1}{1+x}=\frac{1}{1-(-x)}=1-x+x^2-x^3+\cdots \}$

$$\begin{array}{rl}& [\ln (1+x){]}^{\prime }=\frac{1}{1+x}\\ \Rightarrow & \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \\ \Rightarrow & \ln (1+x^2)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots \\ & \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots \\ \Rightarrow & 1-\cos x=\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}-\cdots \end{array}$$

4 (Continued)

$$\underset{x\to 0}{lim}\frac{\ln (1+x^2)}{1-\cos x}=\underset{x\to 0}{lim}\frac{x^2-\frac{x^4}{2}+\frac{x^6}{3}-\cdots }{\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}\cdots }=2$$

5. $L={\int }_0^{2\pi }\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}={\int }_0^{2\pi }\frac{3}{2}|\sin 2t|dt=(4)\left(\frac{3}{2}\right){\int }_0^{\frac{\pi }{2}}\sin 2tdt$ $=-3\cos 2t{]}_0^{\frac{\pi }{2}}=6$

6. $r=2\sin \theta ,\frac{\pi }{4}⩽\theta ⩽\frac{\pi }{2}$

$r^2=2r\sin \theta ,x^2+y^2=2y,x^2+(y-1{)}^2=1$

$$\begin{array}{rl}A& =\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{r}^{2}d\theta ={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}2{\sin }^2\theta d\theta \\ & ={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}[1-\cos 2\theta ]d\theta ={[\theta -\frac{\sin 2\theta }{2}]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}=\frac{\pi }{4}+\frac{1}{2}\end{array}$$

Alternative method:

$$\begin{array}{rl}A& =\frac{1}{4}(\text{ area of disc })+\text{ area of the triangle }\\ & =\frac{\pi }{4}+\frac{1}{2}\end{array}$$

7. $(1+x{)}^m=1+\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{k})x^k,\left(\frac{m}{k}\right)=\frac{m(m-1)(m-2)\cdots (m-k+1)}{k!}$, for $|x|<1$

$(1+x{)}^{\frac{1}{2}}=1+\frac{1}{2}x-\frac{1}{8}{x}^2+\frac{1}{16}{x}^3+\cdots$ for $|x|<1$

8. First assume $\left\{x_n\right\}$ conv. to 1 . Then $\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}{x}_n=L$

Thus $L=\frac{L}{2}+\frac{1}{2}\Rightarrow L=\sqrt{2}$ (we observe $L\ge 0$ )

Now we show $\left\{x_n\right\}$ cons. indeed:

• Since $x_n>0$, we see $x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}⩾\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$ for all $n$

• $x_{n+1}-x_n=\frac{2-x_n{}^2}{2x_n}⩽0$

Thus, $\left\{x_n\right\}$ is nonincreasing and bounded from below by $\frac{\sqrt{2}}{2}$, and therefore it must converge