10.6.44 Determine whether the series converges absolutely, conditionally or diverges.

[karush]

$\tiny{10.6.44}\\$
$\textsf{Does $S_n$ Determine whether the series converges absolutely, conditionally or diverges.?}\\$
\begin{align*}\displaystyle
S_n&= \sum_{n=1}^{\infty}
\frac{(-1)^n}{\sqrt{n}+\sqrt{n+6}}\\
\end{align*}
$\textit {apparently the ratio and root tests fail}$

 

[Prove It]

$\tiny{10.6.44}\\$
$\textsf{Does $S_n$ Determine whether the series converges absolutely, conditionally or diverges.?}\\$
\begin{align*}\displaystyle
S_n&= \sum_{n=1}^{\infty}
\frac{(-1)^n}{\sqrt{n}+\sqrt{n+6}}\\
\end{align*}
$\textit {apparently the ratio and root tests fail}$

Well to determine absolute convergence, we need to first look at the absolute value series:

$\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{1}{\sqrt{n} + \sqrt{n + 6}} } \end{align*}$

Since $\displaystyle \begin{align*} \sqrt{n} < \sqrt{n + 6} \end{align*}$ that means $\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{1}{\sqrt{n} + \sqrt{n + 6}} } > \sum_{n = 1}^{\infty}{ \frac{1}{\sqrt{n+ 6} + \sqrt{n + 6}} } \end{align*}$, now notice that

$\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{\frac{1}{\sqrt{n+6} + \sqrt{n+6}} } &= \sum_{n = 1}^{\infty}{ \frac{1}{2\,\sqrt{n+6}} } \\ &= \frac{1}{2}\,\sum_{n = 1}^{\infty}{ \frac{1}{\sqrt{n+6}} } \\ &= \frac{1}{2}\,\sum_{n = 1}^{\infty}{ \frac{1}{\left( n + 6 \right) ^{\frac{1}{2}}} } \end{align*}$

and this is a divergent p-series, so the absolute value series diverges by comparison.

Thus our original series is NOT absolutely convergent.

As for testing conditional convergence, it's an alternating series, so try the alternating series test.

 

[karush]

so that is what that means!