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\begin{document}
\begin{center}
{\bf \Large Math 112 Homework 5 Solutions}\medskip
\end{center}

{\bf Problem 1.}

Since $(\sqrt{a_n}-\frac{1}{n})^2=a_n+\frac{1}{n^2}-\frac{2}{n}\sqrt{a_n}$,
we have $$0\le \frac{1}{n}\sqrt{a_n}\le \frac{1}{2}(a_n+\frac{1}{n^2}).$$
The series $\sum a_n$ converges by assumption, and the series $\sum
\frac{1}{n^2}$ converges, so by the comparison criterion the series $\sum
\frac{1}{n}\sqrt{a_n}$ converges.

\medskip

{\bf Problem 2.}

a) $\lim\sup_{n\to\infty} \sqrt[n]{n^3}=\lim\sup_{n\to\infty}n^{3/n}=
\lim\sup_{n\to\infty} (n^{1/n})^3=1$, so $R=1$.

b) Note that $\sum\limits_{n=0}^\infty \dfrac{2^n}{n!}
z^n=\sum\limits_{n=0}^\infty \dfrac{(2z)^n}{n!}=e^{2z}$.
We know that $e^z$ has $R=\infty$, i.e.\ $\lim\sup(n!)^{-1/n}=0$ (see
Example 3.40(b)). Therefore $\lim\sup (2^n/n!)^{1/n}=2\lim\sup
(1/n!)^{1/n}=0$, and $R=\infty$ again.

c) Observe that, for any $k\in\Z$, $\lim_{n\to\infty}(n^k)^{1/n}=1$.
This is because $\lim n^{1/n}=1$ (Theorem 3.20(c)), so $\lim_{n\to\infty}
n^{k/n}=1$. Hence, we have $\lim\sup_{n\to\infty} (2^n/n^2)^{1/n}=
2\lim\sup_{n\to\infty} n^{-2/n}=2$, so $R=1/2$.

d) Similarly, $\lim\sup_{n\to\infty} (n^3/3^n)^{1/n}=1/3$, so $R=3$.
\medskip

{\bf Problem 3.}

a) Observe that, if $a_n\ge 1$, then $\frac{a_n}{1+a_n}\ge
\frac{a_n}{a_n+a_n}=\frac{1}{2}$, while if $a_n\le 1$, then 
$\frac{a_n}{1+a_n}\ge \frac{a_n}{1+1}=\frac{a_n}{2}$. 
We consider the following alternative.

Case 1: there exist infinitely many values of $n$ such that $a_n>1$.
Then the corresponding values of $\frac{a_n}{1+a_n}$ are all greater than
$\frac{1}{2}$, so $\frac{a_n}{1+a_n}\not\to 0$, and therefore
$\sum\frac{a_n}{1+a_n}$ diverges.

Case 2: for all but finitely many values of $n$, we have $a_n\le 1$ and 
therefore $\frac{a_n}{1+a_n}\ge\frac{a_n}{2}$. Since a finite number of
terms cannot affect the behavior of the series, by the
comparison criterion we can conclude that, since $\sum\frac{a_n}{2}$ is
divergent, the series $\sum \frac{a_n}{1+a_n}$ is also divergent.

(Shorter alternate solution: if $a_n\not\to 0$ then $\frac{a_n}{1+a_n}
\not\to 0$ and then $\sum \frac{a_n}{1+a_n}$ diverges. If $a_n\to 0$
then there exists $N$ such that, for $n\ge N$, $a_n\le 1$ and therefore
$\frac{a_n}{1+a_n}\ge \frac{a_n}{2}$ $\forall n\ge N$, which gives
divergence by the comparison criterion).

b) Since $a_n>0$, the sequence $\{s_n\}$ is monotonically increasing,
and therefore $\frac{a_{N+1}}{s_{N+1}}+\dots+\frac{a_{N+k}}{s_{N+k}}\ge
\frac{a_{N+1}}{s_{N+k}}+\dots+\frac{a_{N+k}}{s_{N+k}}=\frac{a_{N+1}+\dots+
a_{N+k}}{s_{N+k}}=\frac{s_{N+k}-s_N}{s_{N+k}}=1-\frac{s_N}{s_{N+k}}$.

Observe that the divergence of the series $\sum a_n$ implies that the
monotonically increasing sequence $s_n$ is not bounded (by Theorem 3.24).
In particular, given any integer $N$, the constant $2s_N$ is not an upper 
bound for $\{s_n\}$, i.e.\ there exists $m$ such that $s_m\ge 2s_N$; 
obviously $m>N$, so we can write $m=N+k$
for some positive integer $k$. Therefore, for every integer $N$, there
exists an integer $k$ such that $s_{N+k}\ge 2s_N$, i.e.\
$1-\frac{s_N}{s_{N+k}}\ge \frac{1}{2}$. As a consequence, we obtain:
$\forall N$ $\exists k$ such that $\sum_{n=N+1}^{N+k} \frac{a_n}{s_n}\ge 
\frac{1}{2}$.

Assume that the series $\sum\frac{a_n}{s_n}$ is convergent: then by the
Cauchy criterion (Theorem 3.22) there exists $N$ such that, $\forall m\ge
n\ge N$, $|\sum_{k=n}^m \frac{a_k}{s_k}|\le \frac{1}{3}$. This contradicts
the previously obtained statement; therefore $\sum\frac{a_n}{s_n}$ diverges.

c) Since $s_n=s_{n-1}+a_n\ge s_{n-1}$, we have
$\frac{1}{s_{n-1}}-\frac{1}{s_n}=\frac{s_n-s_{n-1}}{s_{n-1}s_n}=
\frac{a_n}{s_{n-1}s_n}\ge \frac{a_n}{s_n^2}$. Therefore $\sum_{n=1}^N
\frac{a_n}{s_n^2}\le \frac{a_1}{s_1^2}+\sum_{n=2}^N (\frac{1}{s_{n-1}}-
\frac{1}{s_n})=\frac{1}{a_1}+(\frac{1}{s_1}-\frac{1}{s_2})+\dots+
(\frac{1}{s_{N-1}}-\frac{1}{s_N})=\frac{2}{a_1}-\frac{1}{s_N}<\frac{2}{a_1}$.
So the partial sums of the series of non-negative terms
$\sum\frac{a_n}{s_n^2}$ are bounded by the constant $\frac{2}{a_1}$, and
therefore by Theorem 3.24 the series $\sum\frac{a_n}{s_n^2}$ is convergent.

d) The series $\sum\frac{a_n}{1+na_n}$ can be either divergent or
convergent. Divergent examples are easy to come by (e.g., if $a_n=1$ then
$\frac{a_n}{1+na_n}=\frac{1}{n+1}$ and $\sum\frac{1}{n+1}$ is divergent;
if $a_n=\frac{1}{n}$ then $\frac{a_n}{1+na_n}=\frac{1}{2n}$ and
$\sum\frac{1}{2n}$ is divergent). 

An example where $\sum a_n$ diverges
but $\sum\frac{a_n}{1+na_n}$ converges is given by setting
$a_n=1$ if there exists $k$ such that $n=k^2$, and $a_n=\frac{1}{n^2}$
otherwise. Because infinitely many terms are equal to $1$ the sequence
$\{a_n\}$ does not converge to $0$; therefore $\sum a_n$ is divergent.
Meanwhile, if $n=k^2$ then $\frac{a_n}{1+na_n}=\frac{1}{k^2+1}<\frac{1}{k^2}$,
while in the other case $\frac{a_n}{1+na_n}<a_n=\frac{1}{n^2}$. Therefore
the partial sums of the series $\sum \frac{a_n}{1+na_n}$ are bounded:
$\sum_{n=1}^N \frac{a_n}{1+na_n}\le \sum_{k^2\le N} \frac{1}{k^2}+
\sum_{n=1}^N \frac{1}{n^2}\le \sum_{k=1}^\infty \frac{1}{k^2}+\sum_{n=1}^\infty 
\frac{1}{n^2}$ (recall $\sum \frac{1}{n^2}$ is convergent). By Theorem 3.24 we
conclude that in this example $\sum \frac{a_n}{1+na_n}$ is convergent.

The case of the series $\sum\frac{a_n}{1+n^2a_n}$ is easier: it is always
convergent, by the comparison criterion, because $0\le \frac{a_n}{1+n^2a_n}
\le\frac{1}{n^2}$.

\end{document}
