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\begin{center}Math 132: Calculus for Physical and Life Sciences 2 \\ Problem Set 8 \\ Due Friday, April 11, 2008, at the beginning of class.\\ Solutions\end{center}\vspace{.1in}



\begin{enumerate}
\item Determine whether each of the following sequences converges or diverges. If it converges, find its limit. \vspace{.1in}



(a) $\displaystyle a_n=\frac{1-n^2}{2+3n^2}$ \vspace{.1in}

\emph{Solution:} $\displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2}{2+3n^2}=  \lim_{n \rightarrow \infty} \frac{\frac{1}{n^2}-1}{\frac{2}{n^2}+3}=- \frac{1}{3}.$\vspace{.1in}

Or use  l'Hospital's rule twice: $\displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2}{2+3n^2}= \lim_{x \rightarrow \infty} \frac{1-x^2}{2+3x^2}\stackrel{l'H}{=} \lim_{x \rightarrow \infty} \frac{-2x}{6x}\stackrel{l'H}{=} \lim_{x \rightarrow \infty} \frac{-2}{6}=-\frac{1}{3}$\vspace{.1in}

(b)  $\displaystyle a_n=1+(-1)^n$\vspace{.1in}

\emph{Solution:} $a_n=0$ if $n$ is odd and $a_n=2$ if $n$ is even. The sequence does not approach a single number and thus it diverges. \vspace{.1in}


(c) $\displaystyle a_n= \frac{\sin n}{3^n}$\vspace{.1in}

\emph{Solution:} Use the Squeeze Theorem. $-1 \leq \sin n \leq 1$ for all $n$. Thus $\displaystyle - \frac{1}{3^n} \leq \frac{\sin n}{3^n} \leq \frac{1}{3^n}$. Since $\displaystyle \lim_{n \rightarrow \infty} -\frac{1}{3^n}=\lim_{n \rightarrow \infty} \frac{1}{3^n}=0$, we have that $\displaystyle \lim_{n \rightarrow \infty}\frac{\sin n}{3^n}=0$.
\vspace{.1in}


(d) $\displaystyle a_n=\frac{(\ln n)^2}{n}$\vspace{.1in}

\emph{Solution:} $\displaystyle \lim_{n \rightarrow \infty}\frac{(\ln n)^2}{n}= \lim_{x \rightarrow \infty}\frac{(\ln x)^2}{x}\stackrel{l'H}{=}\lim_{x \rightarrow \infty}\frac{2(\ln x)\frac{1}{x}}{1}= \lim_{x \rightarrow \infty}\frac{2\ln x}{x}\stackrel{l'H}{=} \lim_{x \rightarrow \infty}\frac{2}{x}=0$.\vspace{.1in}

(e) $\displaystyle a_n=\frac{2^n+1}{e^n}$\vspace{.1in}

\emph{Solution:} $\displaystyle \lim_{n \rightarrow \infty}\frac{2^n+1}{e^n}=  \lim_{n \rightarrow \infty}\left(\frac{2}{e}\right)^n +  \lim_{n \rightarrow \infty}\left(\frac{1}{e}\right)^n =0$ since both $\displaystyle \frac{2}{e}$ and $\displaystyle \frac{1}{e}$ are between $0$ and $1$. \vspace{.1in}


(f) $\displaystyle a_n=\ln(2n+1)-\ln(2n-1)$ \vspace{.1in}

\emph{Solution:} $\displaystyle \lim_{n \rightarrow \infty}a_n=  \lim_{n \rightarrow \infty} \ln \left( \frac{2n+1}{2n-1}\right)= \ln \left(\lim_{n \rightarrow \infty} \frac{2n+1}{2n-1}\right)=0$.\vspace{.1in}


\item Determine if each infinite series converges or diverges. If it converges, find its sum. (Be careful about the first term of the series.)\vspace{.1in}

\ (a) $1+ e^{-1} + e^{-2}+e^{-3}+ \ldots + e^{-n} + \ldots$\vspace{.1in}

\emph{Solution:} $\displaystyle \sum_{n=0}^{\infty} (e^{-1})^n$ is the geometric series with ratio $\displaystyle \frac{1}{e}$ and first term $1$. It is convergent and its sum is $\displaystyle\frac{e}{e-1}$. \vspace{.1in}


(b) $\displaystyle \sum_{n=0}^{\infty}\frac{3^n-2^n}{4^n}$\vspace{.1in}

\emph{Solution:} $\displaystyle \sum_{n=0}^{\infty} \frac{3^n}{4^n}=  \sum_{n=0}^{\infty} \left(\frac{3}{4}\right)^n= 4$ (geometric series with ratio $3/4$ and first term $1$).   $\displaystyle  \sum_{n=0}^{\infty} \left(\frac{2}{4}\right)^n= 2$ (geometric series with ratio $1/2$ and first term $1$). 

Thus $\displaystyle \sum_{n=0}^{\infty}\frac{3^n-2^n}{4^n}= \sum_{n=0}^{\infty} \frac{3^n}{4^n}- \sum_{n=0}^{\infty} \frac{2^n}{4^n}= 4-2=2$\vspace{.1in}

(c) $\displaystyle \sum_{n=0}^{\infty} \left( \frac{100}{99}\right) ^n$\vspace{.1in}

\emph{Solution:} This is a geometric series with ratio $100/99>1$. It diverges. \vspace{.1in}

(d) $\displaystyle \sum_{n=0}^{\infty} \left( \frac{99}{100}\right) ^n$ \vspace{.1in}

\emph{Solution:} This is a geometric series with ratio $99/100<1$ and first term $1$. It converges to $100$. \vspace{.1in}


(e) $\displaystyle \sum_{n=1}^{\infty} \left( \frac{\pi}{e}\right) ^n$\vspace{.1in}


\emph{Solution:} This is a geometric series with ratio $\pi/e>1$. It diverges. \vspace{.1in}

 (f) $\displaystyle \sum_{n=1}^{\infty} \left( \frac{2}{n}-\frac{1}{2^n}\right) $\vspace{.1in}
 
 \emph{Solution:} $\displaystyle \sum_{n=0}^{\infty}\frac{2}{n}= 2 \displaystyle \sum_{n=0}^{\infty}\frac{1}{n}$ is divergent (twice the harmonic series). $\displaystyle \sum_{n=0}^{\infty}\frac{1}{2^n}$ converges (geometric series with ratio $1/2$). Thus $\displaystyle \sum_{n=1}^{\infty} \left( \frac{2}{n}-\frac{1}{2^n}\right) $ diverges. ( If $\displaystyle \sum_{n=1}^{\infty} \left( \frac{2}{n}-\frac{1}{2^n}\right) $ were convergent, $\displaystyle \sum_{n=0}^{\infty}\frac{2}{n}=\displaystyle \sum_{n=1}^{\infty} \left( \frac{2}{n}-\frac{1}{2^n}\right) + \displaystyle \sum_{n=0}^{\infty}\frac{1}{2^n}$ would be convergent as the sum of two convergent series.)

 
  (g) $\displaystyle \sum_{k=1}^{\infty} \left( \frac{k+1}{2k+4}\right)$\vspace{.1in}

\emph{Solution:} Since $\displaystyle \lim_{k \rightarrow \infty}  \frac{k+1}{2k+4}= \frac{1}{2} \neq 0$, the series diverges by the Divergence Test. \vspace{.1in}

\ (h) $\displaystyle \sum_{n=1}^{\infty} \ln\left( \frac{n+1}{n}\right)$ \vspace{.1in}

(Hint: Consider several partial sums and use properties of logarithms to simplify them.)\vspace{.1in}

\emph{Solution:}  Use the fact that $ \displaystyle \ln\left( \frac{n+1}{n}\right)= \ln(n+1)-\ln n$. Consider the $n$-th partial sum 

$s_n= (\ln 2 - \ln 1) + (\ln 3 - \ln 2) + (\ln 4 - \ln 3) + \cdots +(\ln n - \ln(n-1))+ (\ln(n+1)-\ln n)$. All terms cancel except for $\ln 1$ and $\ln(n+1)$. Thus $s_n=\ln(n+1)-\ln 1=\ln(n+1)$. Since $\displaystyle \lim_{n\rightarrow \infty} s_n = \infty$, the series diverges. \vspace{.1in}

\item Find the rational number represented by  $0.2525252525\ldots$. 

(Hint: Think of $0.2525252525\ldots$ as a geometric series.)  \vspace{.1in}

\emph{Solution:} $\displaystyle 0.2525252525\ldots= \frac{25}{100}+ \frac{25}{(100)^2}+  \frac{25}{(100)^3}+ \cdots$ This is the geometric series with ratio $1/100$ and first term $25/100$. It is convergent and its sum is $$ \frac{\frac{25}{100}}{1-\frac{1}{100}}=\frac{25}{99}.$$ Thus $\displaystyle 0.2525252525\ldots=\frac{25}{99}$. \vspace{.1in}

\item Suppose the government spends \$1 billion and that each recipient spends $90\%$ of the dollars that he or she receives. In turn, the secondary recipients spend $90\%$ of the dollars they receive, and so on. How much total spending results from the original injection of  \$1 billion into the country? \vspace{.1in}

\emph{Solution:} The total spending (in Dollars) equals $\displaystyle 10^9+10^9 \cdot 0.9+ 10^9(0.9)^2+ 10^9(0.9)^3 + \cdots = \frac{10^9 }{1-0.9}= 9 \cdot 10^9$ since this is a geometric series with ratio $0.9$ and first term $10^9$. Thus the total spending is $\$10$ billion.  \vspace{.1in}

\item Use the integral test to decide whether each of the following series converges or diverges.  \vspace{.1in}



(a) $\displaystyle \sum_{n=0}^{\infty}\frac{n}{n^2+1}$ \vspace{.1in}

\emph{Solution:} Let $\displaystyle f(x)=\frac{x}{x^2+1}$. By the  quotient rule (check!) $\displaystyle f'(x)= \frac{1-x^2}{(x^2+1)^2}$. For $x >1$ this is negative and thus $f$ is decreasing. The function $f$ is also continuous and positive for $x>0$. We can apply the integral test (use the $u$-substitution $u=x^2+1,\ du=2 x \, dx$ in the integral). 

$$ \int_1^{\infty} \frac{x}{x^2+1}\, dx=\lim_{b \rightarrow \infty} \int_1^b \frac{x}{x^2+1}\, dx = \lim_{b \rightarrow \infty}\frac{1}{2} \int_2^{b^2+1} \frac{1}{u}\, du =  \lim_{b \rightarrow \infty}\frac{1}{2}(\ln(b^2+1)-\ln 2)=\infty.$$ Since the integral is divergent, the series is divergent as well. \vspace{.1in}

(b) $\displaystyle \sum_{n=0}^{\infty} \frac{n}{e^{n^2}}$ \vspace{.1in}

\emph{Solution:} Let $\displaystyle f(x)= \frac{x}{e^{x^2}}$. We have $\displaystyle f'(x)= \frac{e^{x^2}-2x^2 e^{x^2}}{(e^{x^2})^2}= \frac{e^{x^2}(1-2x^2)}{e^{2x^2}}.$ Since $f'(x)<0$ for $x\geq 1$, the function $f$ is decreasing. The function $f$ is also continuous and positive for $x>0$ and we can apply the integral test (use the $u$-substitution $u=x^2,\ du=2 x \, dx$ in the integral). $$\int_1^{\infty}  \frac{x}{e^{x^2}}\, dx = \lim_{b \rightarrow \infty} \int_1^b  \frac{x}{e^{x^2}}\, dx =  \lim_{b \rightarrow \infty} \frac{1}{2}\int_1^{b^2} e^{-u}\, du =  \lim_{b \rightarrow \infty} \frac{1}{2}(-e^{-b^2}+ e^{-1}) =\frac{1}{2} e^{-1}=\frac{1}{2e}$$ Since the integral is convergent, the series converges as well. \vspace{.1in}

(c) $\displaystyle \sum_{n=2}^{\infty}\frac{1}{n \ln n}$ \vspace{.1in}

\emph{Solution:} Let $\displaystyle f(x)=\frac{1}{x \ln x}$. Since the denominator is increasing, the function is decreasing. It is positive and continuous for $x>1$ and we can apply the integral test (use the $u$-substitution $u=\ln x,\ du=1/x \, dx$ in the integral). $$\int_2^{\infty}  \frac{1}{x \ln x}\, dx = \lim_{b \rightarrow \infty} \int_2^b \frac{1}{x \ln x}\, dx = \lim_{b \rightarrow \infty} \int_{\ln 2}^{\ln b} \frac{1}{u}\, du=  \lim_{b \rightarrow \infty} (\ln(\ln b)-\ln(\ln 2))= \infty.$$ Since the integral diverges, the series diverges as well. \vspace{.1in}


(d)  $\displaystyle \sum_{n=1}^{\infty}\frac{n}{e^n}$ \vspace{.1in}

\emph{Solution:} Let $\displaystyle f(x)=\frac{x}{e^x}$. Since  $\displaystyle f'(x)=\frac{e^x(1-x)}{e^{2x}}$ is negative for $x>1$, the function $f$ is 

decreasing for $x>1$. Also, for $x>0$, $f$ is continuous and positive. We can apply the integral test (use integration by parts in the definite integral: $u=x\, du= 1 \, dx$ and $dv=e^{-x}\, dx, v=-e^{-x}$). $$\int_1^{\infty} \frac{x}{e^x}\, dx= \lim_{b \rightarrow \infty} \int_1^b x e^{-x}\, dx =  \lim_{b \rightarrow \infty} (-xe^{-x}|_1^b + \int_1^b e^{-x}\, dx=  \lim_{b \rightarrow \infty} (-be^{-b}-e^{-b} + 2 e^{-1}).$$ By l'Hospital's rule $ \displaystyle \lim_{b \rightarrow \infty} be^{-b}=  \lim_{b \rightarrow \infty} \frac{b}{e^{b}}= \lim_{b \rightarrow \infty}\frac{1}{e^b}=0$. Thus $\displaystyle \int_1^{\infty} \frac{x}{e^x}\, dx = 2 e^{-1}$. Since the integral is convergent, the series is convergent as well. \vspace{.1in}

(e) $\displaystyle \sum_{n=1}^{\infty}\ln \left( 1 + \frac{1}{n^2}\right)$ \vspace{.1in}

(Hint: integrate by parts.) \vspace{.1in}

\emph{Solution:} Let $\displaystyle f(x)=\ln \left( 1 + \frac{1}{x^2}\right)$. It is a positive, continuous and decreasing function for $x \geq 1$ and we can apply the integral test. We will first solve $\displaystyle \int \ln \left( 1 + \frac{1}{x^2}\right)\, dx$ using integration by parts. Let $\displaystyle  u=\ln \left( 1 + \frac{1}{x^2}\right)$ and $dv = dx$. Then $\displaystyle du= \frac{1}{1+\frac{1}{x^2}}\cdot \frac{-2}{x^3}\, dx= -\frac{2}{x(x^2+1)}\, dx$ and $v=x$. Then $\displaystyle \int \ln \left( 1 + \frac{1}{x^2}\right)\, dx= x\ln \left( 1 + \frac{1}{x^2}\right) + 2 \int \frac{1}{1+x^2}\, dx = x\ln \left( 1 + \frac{1}{x^2}\right)  + 2 \arctan x + C.$ Then $$ \int_1^{\infty} \ln \left( 1 + \frac{1}{x^2}\right)\, dx=  \lim_{b \rightarrow \infty} \int_1^b \ln \left( 1 + \frac{1}{x^2}\right)\, dx=  \lim_{b \rightarrow \infty} \left(b \ln \left( 1 + \frac{1}{b^2}\right)  + 2 \arctan b - \ln 2+ 2 \arctan 2\right)$$

We have $\displaystyle  \lim_{b \rightarrow \infty} b \ln \left( 1 + \frac{1}{b^2}\right) =  \lim_{b \rightarrow \infty} \frac{ \ln \left( 1 + \frac{1}{b^2}\right)}{\frac{1}{b}} \stackrel{l'H}{=} \lim_{b \rightarrow \infty}\frac{\frac{1}{1+\frac{1}{b^2}}\cdot \frac{-2}{b^3}}{-\frac{1}{b^2}} =\lim_{b \rightarrow \infty}\frac{2}{b(1+\frac{1}{b^2})}=0$ and $\lim_{b \rightarrow \infty}\arctan b= \pi/2$.

Therefore the integral converges and so does the series. 




\item Determine whether the given series converges or diverges. Quote general results to justify your answers.  \vspace{.1in}


(a)  $\displaystyle \sum_{n=1}^{\infty}\frac{1}{(\sqrt{n})^5}$ \vspace{.1in}

\emph{Solution:}  $\displaystyle \sum_{n=1}^{\infty}\frac{1}{(\sqrt{n})^5}=  \sum_{n=1}^{\infty}\frac{1}{n^{5/2}}$. This is the $p$-series with $p=5/2>1$ and is therefore convergent. \vspace{.1in}

(b)  $\displaystyle \sum_{n=1}^{\infty}\frac{n+1}{n^2}$ \vspace{.1in}


\emph{Solution:} Since $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ diverges (harmonic series) and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ convergent ($p$-series with $p=2$), the given series diverges.  \vspace{.1in}


\item Show that the series $ \displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$ converges if $p>1$ and diverges if $p\leq1$. \vspace{.1in}

\emph{Solution:} If $p=1$ this is the series of problem 5.c) which diverges. Suppose $p \neq 1$. Since $\displaystyle f(x)=\frac{1}{x(\ln x)^p}$ is positive, continuous and decreasing for $x>1$ we use the integral test. In the integral we use the substitution $u=\ln x$, $du=1/x\, dx$. 

$$  \int_2^{\infty} \frac{1}{x(\ln x)^p}\, dx= \lim_{b \rightarrow \infty} \int_2^b \frac{1}{x(\ln x)^p}\, dx= \lim_{b \rightarrow \infty} \int_{\ln 2}^{\ln b}\frac{1}{u^p}\, du=  \lim_{b \rightarrow \infty} \left.\frac{ u^{-p+1}}{-p+1}\right|_{\ln 2}^{\ln b}=$$ $$ = \lim_{b \rightarrow \infty} \left(\frac{ (\ln b)^{-p+1}}{-p+1}-\frac{ (\ln 2)^{-p+1}}{-p+1}\right)$$

If $p>1$, then $\displaystyle  \lim_{b \rightarrow \infty}  (\ln b)^{-p+1}=0$ and if $p<1$, then $\displaystyle  \lim_{b \rightarrow \infty}  (\ln b)^{-p+1}=\infty$. Therefore, the integral converges is $p>1$ and it diverges if $p<1$. By the integral test and problem 5.c), the series converges if $p>1$ and it diverges if $p\leq 1$. \vspace{.1in} 

\item For each of the following series explain why the integral test does \emph{not} apply. 

\vspace{.1in} 

(a) $ \displaystyle \sum_{n=1}^{\infty} e^{-n}\sin{n}$ \vspace{.1in} 

\emph{Solution:} The integral test does not apply because the function $f(x)=e^{-x}\sin{x}$ is not positive (for example it takes negative values at each odd multiple of $\pi/2$) and not decreasing ($f'(x)=e^{-x}(\cos x - \sin x)$ is not always negative - for example, it is positive at all even multiples of $\pi$).  \vspace{.1in} 


b) $ \displaystyle \sum_{n=1}^{\infty}\frac{2+\sin n}{n^2}$\vspace{.1in} 

\emph{Solution:} Here the function $\displaystyle f(x) = \frac{2+\sin x}{x^2}$ is positive for all $x\neq 0$ (because $-1 \leq \sin x \leq 1$). However, the function is not decreasing since $\displaystyle f'(x) =\frac{x^2 \cos x - 2x(2+\sin x)}{x^4}$ is not always negative (it is positive for all even multiples of $\pi$ greater than $4$). 


\end{enumerate}\vspace{.2in}



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