MHB Does the Series Converge Absolutely, Conditionally, or Diverge?

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$\tiny{10.6.44}\\$
$\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*}
$\textit {apparently the ratio and root tests fail}$
 
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karush said:
$\tiny{10.6.44}\\$
$\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.
 
so that is what that means!
 
Last edited:

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