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\noindent
\parbox{1.5in}{\bf Math 1120}
\hfill {\Large \bf Series Tests} \hfill
\parbox{1.5in}{\bf \hfill Mixed Practice}

\vspace{0.25in}

\begin{minipage}[t]{2.25in}
\begin{description}
\item[\#1.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^2+n}$
\item[\#2.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{n-1}{n^2+n}$
\item[\#3.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{1}{n^2+n}$
\item[\#4.] \quad $\displaystyle\sum\limits_{n=1}^{\infty}(-1)^{n-1} \frac{n-1}{n^2+n}$
\item[\#5.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{(-3)^{n+1}}{2^{3 n}}$
\item[\#6.] \quad $\displaystyle\sum\limits_{n=1}^{\infty}\left(\frac{3 n}{1+8 n}\right)^n$
\item[\#7.] \quad $\displaystyle\sum\limits_{n=2}^{\infty} \frac{1}{n \sqrt{\ln (n)}}$
\item[\#8.] \quad $\displaystyle\sum\limits_{k=1}^{\infty} \frac{2^k k !}{(k+2) !}$
\item[\#9.] \quad $\displaystyle\sum\limits_{k=1}^{\infty} k^2 e^{-k}$
\item[\#10.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} n^2 e^{-n^3}$
\item[\#11.] \quad $\displaystyle\sum\limits_{n=2}^{\infty} \frac{(-1)^{n+1}}{n \ln (n)}$
\item[\#12.] \quad $\displaystyle\sum\limits_{n=1}^{\infty}(-1)^n \frac{n}{n^2+25}$
\item[\#13.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{3^n n^2}{n !}$
\end{description}
\end{minipage} \hfill
\begin{minipage}[t]{2.25in}
\begin{description}
\item[\#14.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \sin (n)$
\item[\#15.] \quad $\displaystyle\sum\limits_{n=0}^{\infty} \frac{n !}{2 \cdot 5 \cdot 8 \cdots \cdots(3 n+2)}$
\item[\#16.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{n^2+1}{n^3+1}$
\item[\#17.] \quad $\displaystyle\sum\limits_{n=1}^{\infty}(-1)^n 2^{1 / n}$
\item[\#18.] \quad $\displaystyle\sum\limits_{n=2}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}-1}$
\item[\#19.] \quad $\displaystyle\sum\limits_{n=1}^{\infty}(-1)^n \frac{\ln (n)}{\sqrt{n}}$
\item[\#20.] \quad $\displaystyle\sum\limits_{k=1}^{\infty} \frac{k+5}{5^k}$
\item[\#21.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{(-2)^{2 n}}{n^n}$
\item[\#22.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{\sqrt{n^2-1}}{n^3+2 n^2+5}$
\item[\#23.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \tan (1 / n)$
\item[\#24.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{\cos (n / 2)}{n^2+4 n}$
\item[\#25.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{n !}{e^{n^2}}$
\item[\#26.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{n^2+1}{5^n}$
\end{description}
\end{minipage} \hfill
\begin{minipage}[t]{2.25in}
\begin{description}
\item[\#27.] \quad $\displaystyle\sum\limits_{k=1}^{\infty} \frac{k \ln (k)}{(k+1)^3}$
\item[\#28.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{e^{1 / n}}{n^2}$
\item[\#29.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{\tan ^{-1}(n)}{n \sqrt{n}}$
\item[\#30.] \quad $\displaystyle\sum\limits_{j=1}^{\infty}(-1)^j \frac{\sqrt{j}}{j+5}$
\item[\#31.] \quad $\displaystyle\sum\limits_{k=1}^{\infty} \frac{5^k}{3^k+4^k}$
\item[\#32.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{(2 n)^n}{n^{2 n}}$
\item[\#33.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{\sin (1 / n)}{\sqrt{n}}$
\item[\#34.] \quad $\displaystyle\sum\limits_{n=1}^{\infty} \frac{1}{n+n \cos ^2(n)}$
\item[\#35.] \quad $\displaystyle\sum\limits_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^2}$
\item[\#36.] \quad $\displaystyle\sum\limits_{n=2}^{\infty} \frac{1}{(\ln (n))^{\ln (n)}}$
\item[\#37.] \quad $\displaystyle\sum\limits_{n=1}^{\infty}(\sqrt[n]{2}-1)^n$
\item[\#38.] \quad $\displaystyle\sum\limits_{n=1}^{\infty}(\sqrt[n]{2}-1)$
\end{description}
\end{minipage}

\vspace{0.25in}

\newpage

\noindent {\bf Answers with suggested tests:} There are a few places where you might be able to avoid the integral test by using a ratio test or a clever comparison. In other words, I do not claim these solutions are optimal. For conditionally convergent series the suggested tests are given for the {\it original series} (converges) \underline{plus} {\it related absolute values series} (diverges). 

\vspace{0.1in}

\noindent {\bf Abbreviations:} Div. = Diverges, C. Conv. = Conditionally Convergent, Conv. Abs. = Converges Absolutely, 
geom. = geometric series, $p=X$ = $p$-series with exponent $X$, harm. = harmonic series, comp. = comparison (usually limit comparison is intended), dir. comp. = direct comparison, AST = alternating series test, and int. test = integral test.

\vspace{0.15in}

\noindent
\begin{minipage}[t]{2.4in}
\begin{description}
\item[\#1.] Div. ($n^\text{th}$ term)
\item[\#2.] Div. (comp. harm.)
\item[\#3.] Conv. Abs. (comp. $p=2$)
\item[\#4.] C. Conv. (AST plus comp. harm.)
\item[\#5.] Conv. Abs. (geom. r=-3/8)
\item[\#6.] Conv. Abs. (root)
\item[\#7.] Div. (int. test)
\item[\#8.] Conv. Abs. (ratio)
\item[\#9.] Conv. Abs. (ratio)
\item[\#10.] Conv. Abs. (ratio or int test)
\item[\#11.] C. Conv. (AST plus int. test)
\item[\#12.] C.\,Conv.\,(AST\,plus\,comp. harm.)
\item[\#13.] Conv. Abs. (ratio)
\end{description}
\end{minipage} \hfill
\begin{minipage}[t]{2.4in}
\begin{description}
\item[\#14.] Div. ($n^\text{th}$ term)
\item[\#15.] Conv. Abs. (ratio)
\item[\#16.] Div. (comp. harm.)
\item[\#17.] Div. ($n^\text{th}$ term)
\item[\#18.] C. Conv.\\ \force \quad (AST plus comp. $p=0.5$)
\item[\#19.] C. Conv.\\ \force \quad (AST plus dir. comp. $p=0.5$)
\item[\#20.] Conv. Abs. (ratio)
\item[\#21.] Conv. Abs. (root)
\item[\#22.] Conv. Abs. (comp. $p=2$)
\item[\#23.] Div. (comp. harm.)\\ \force \quad [solution below]
\item[\#24.] Conv. Abs. (dir. comp. $p=2$)
\item[\#25.] Conv. Abs. (ratio)
\item[\#26.] Conv. Abs. (ratio)
\end{description}
\end{minipage} \hfill
\begin{minipage}[t]{2.4in}
\begin{description}
\item[\#27.] Conv. Abs. (comp. $p=1.5$)
\item[\#28.] Conv. Abs. (dir. comp. $p=2$)
\item[\#29.] Conv. Abs. (dir. comp. $p=1.5$)\\ \force \quad Note: $\mathrm{arctan}(x) \leq \pi/2$
\item[\#30.] C. Conv.\\ \force \quad (AST plus comp. $p=0.5$)
\item[\#31.] Div. (ratio or $n^\text{th}$ term)
\item[\#32.] Conv. Abs. (root)
\item[\#33.] Conv. Abs. (dir. comp. $p=1.5$)\\ \force \quad [solution below]
\item[\#34.] Div. (dir. comp. harm.)
\item[\#35.] Div. ($n^\text{th}$ term)
\item[\#36.] Conv. Abs. (tricky dir. comp. $p=2$)
\item[\#37.] Conv. Abs. (root)
\item[\#38.] Div. (dir. comp. harm.)\\ \force \quad [solution below]
\end{description}
\end{minipage}

\vspace{0.15in}

\noindent {\bf \#23:} \quad The linearization (i.e., tangent) of $\tan(x)$ centered at $x=0$ is just $x$. So for small $x$'s (i.e., $1/n$ for large $n$), we have ($\tan(x) \approx x$) (i.e., $\tan(1/n) \approx 1/n$). This rather non-obvious observation leads us to compare with the harmonic series. We have $L = \lim\limits_{n \to \infty} \dfrac{\tan(1/n)}{1/n} = \lim\limits_{t \to 0^+} \dfrac{\tan(t)}{t} = \lim\limits_{t \to 0^+} \dfrac{\mathrm{sec}^2(t)}{1} = \mathrm{sec}^2(0)=1$ (using the fact that $n \to \infty$ means $t=1/n \to 0^+$ as well as an application of L'Hopital's rule). Therefore, we get $0<L=1<\infty$, so the limit comparison test says that series \#23 diverges because the harmonic series $\sum 1/n$ diverges. 

\vspace{0.1in}

\noindent {\bf \#36:} \quad A non-obvious observation: $(\ln(n))^{\ln(n)} = e^{\ln\left((\ln(n))^{\ln(n)}\right)} = e^{\ln(n) \cdot \ln(\ln(n))} = e^{\ln(\ln(n)) \cdot \ln(n)} = e^{\ln\left(n^{\ln(\ln(n))}\right)} = n^{\ln(\ln(n))}$. Next, $\ln(\ln(n)) \geq 2$ for large enough $n$ (Why? $\ln(\ln(\infty))=\ln(\infty)=\infty$).  Therefore, for large enough $n$, $n^2 \leq n^{\ln(\ln(n))}$. Thus for such $n$, $\frac{1}{n^{\ln(\ln(n))}} \leq \frac{1}{n^2}$. Therefore, by direct comparison with the convergent $p$-series ($p=2>1$), series \#36 must converge.

\vspace{0.1in}

\noindent {\bf \#38:} \quad This one is tricky. Since $\sqrt[n]{2}-1 = 2^{1/n}-1 \to 2^{1/\infty}-1=2^0-1=0$, the $n^{\text{th}}$ term test does not help. One can show that $2^{1/n} = e^{\ln(2^{1/n})} = e^{\frac{\ln(2)}{n}} > 1+\frac{\ln(2)}{n}$. [This is most easily seen from the exponential function's Maclaurin series -- which we don't know about yet.] Anyway, this means that $2^{1/n}-1>\frac{\ln(2)}{n}$ so series \#38 diverges by direct comparison with ($\ln(2)$ times) the harmonic series.

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