Convergent series??? Determine for each of the following series whether or not it is convergent. You must justify your answer.
1. $\sum_{n=1}^{\infty} \left(\left(n^3+1\right)^{\frac{1}{3}}-n\right)$.
Could I use $a^3-b^3 = (a-b)(a^2+ab+b^2))$?  If so how?
 Using the hint given we get that
\begin{equation*}
\begin{split}
\left(n^3+1\right)^{\frac{1}{3}}-n &= \frac{\left[\left(n^3+1\right)^{\frac{1}{3}}\right]^3-n^3}{\left[\left(n^3+1\right)^{\frac{1}{3}}\right]^2+\left(n^3+1\right)^{\frac{1}{3}}\cdot n+n^2} \\
&= \frac{n^3+1-n^3}{\left(n^3+1\right)^{\frac{2}{3}}+n\left(n^3+1\right)^{\frac{1}{3}}+n^2} \\
&= \frac{1}{\left(n^3+1\right)^{\frac{2}{3}}+n\left(n^3+1\right)^{\frac{1}{3}}+n^2} \\
&\leq \frac{1}{n^2}.
\end{split}
\end{equation*}
But we know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges ($p$-series with $p = 2 > 1$). Thus, by the comparison test $\sum_{n=1}^{\infty} \left(\left(n^3+1\right)^{\frac{1}{3}}-n\right)$ converges as well.
2. $\sum_{n=2}^{\infty} \frac{1}{n\ln{(n)}^3}$.
 The function $f:[2,\infty) \mapsto \mathbb{R}$ defined by $f(x) := \frac{1}{x\ln{(x)}^3}$ is continuous, positive and monotonely decreasing by examining the derivative $f'(x) = -\frac{\ln{(x)}+3}{x^2(\ln{(x)})^4} < 0$ for all $x\geq 2$.
Since $f(n) = \frac{1}{n\ln{(n)}^3}$, to determine whether or not the series converges it suffices to see what happens to $\int_{2}^{\infty} f(x) \; dx$. Doing so, letting $u = \ln{(x)} \Rightarrow du = \frac{dx}{x}$ gives
\begin{equation*}
\begin{split}
\int_{2}^{t} \frac{1}{x\ln{(x)}^3} \; dx &= \lim_{t\to\infty} \int_{\ln{(2)}}^{t} \frac{1}{u^3} \; du \\
&= \lim_{t\to\infty} \left[-\frac{1}{2u^2}\right]_{\ln{(2)}}^{t} \\
&= \frac{1}{2\ln{(2)}^2}
\end{split}
\end{equation*}
which is finite. This integral converges so by the integral test $\sum_{n=2}^{\infty} \frac{1}{n\ln{(n)}^3}$ converges as well.
3. $\sum_{n=1}^{\infty} \frac{\sin{(\pi+n)}}{\sqrt{n}+2^n}$.
 First we note that
\begin{equation*}
\frac{\sin{(\pi+n)}}{\sqrt{n}+2^n} = -\frac{\sin{(n)}}{\sqrt{n}+2^n}.
\end{equation*}
This series is absolutely convergent if and only if
\begin{equation*}
\sum_{n=1}^{\infty} \left|-\frac{\sin{(n)}}{\sqrt{n}+2^n}\right| = \sum_{n=1}^{\infty} \frac{\sin{(n)}}{\sqrt{n}+2^n}
\end{equation*}
converges. We'll prove that this series does converge by using the comparsion test.
Observe that for any $k\geq 1$,
\begin{equation*}
\frac{\sin{(n)}}{\sqrt{n}+2^n}\leq \frac{1}{\sqrt{n}+2^n}\leq \frac{1}{2^n}.
\end{equation*}
But we know that the series $\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ converges as it's just a geometric series with ratio $\frac{1}{2} < 1$. So the comparison test tells us that $\sum_{n=1}^{\infty} \frac{\sin{(\pi+n)}}{\sqrt{n}+2^n}$ converges, and thus the original series is absolutely convergent.

For the first one,
by Bernoulli's inequality,
$(1+x/n)^n \ge 1+x
$
so
$(1+x)^{1/n} \le 1+x/n
$.
Therefore
$\begin{array}\\
(n^m+1)^{1/m}-n
&=n((1+n^{-m})^{1/m}-1)\\
&\le n((1+n^{-m}/m)-1)\\
&= n(n^{-m}/m)\\
&= n^{-m+1}/m\\
\end{array}\\
$
and the sum of this converges if
$m-1 > 1
$
or
$m > 2
$.
