How to determine whether the series converges or diverges $$\begin{align*}
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)}
\end{align*}$$
I apprpach this problem by using the ratio test.
$$ \lim_{k\rightarrow \infty}\frac{\dfrac{(-1)^{k+1}}{2(k+1)+1}}{\dfrac{(-1)^k}{(2k+1)}}$$
which after simplifying gets me $(2k)/(2k+3) + (1)/(2k+3)$
How does one simplify this to $\sqrt{k+1}$ as the answer it is very perplexing?
 A: Hint. The alternating series $\displaystyle \sum_{k=0}^{\infty}(-1)^{k}a_k$ converges provided:
1) $0< a_{k+1} \le a_k$ for $k$ greater than some index $N$.
2) $\lim_{k\to \infty}a_k = 0$
Just take $a_k:=\frac{1}{2k+1}$ here.
A: Trying the ratio test,
$$\left|\frac{(-1)^{k+1}}{2k+3}\Big/\frac{(-1)^k}{2k+1}\right|
  =\left|\frac{2k+1}{2k+3}\right|\to1$$
as $k\to\infty$, so the ratio test fails.
You can prove the series converges by means of the alternating series test.  To do this show that


*

*the terms of the series are alternately positive and negative;  

*the absolute value of the terms tends to zero;  

*the absolute value of the terms is always decreasing, that is,
$$\frac{1}{2k+1}>\frac{1}{2k+3}\ .$$


See if you can fill in the details.
A: There is a much simpler test you can use, namely the Alternating Series Test, http://en.wikipedia.org/wiki/Alternating_series_test
It essentially says that if the terms being added alternate between positive and negative values, and furthermore if the terms approach zero then the sum will converge.
At this point we can ask the following additional question.  Is it that it converged only because it alternated signs?  Or is it that it would have converged even if it didn't alternate signs?
If the answer to that question is that it converges with the alternating positive to negative each time but that it did not converge if you consider the sum of the absolute values, we say it converges conditionally.  If it would have converged even ignoring the $(-1)^k$ piece, in other words if the sum of the absolute values of each term converges, we say it converges absolutely.
In this case, it converges conditionally, since the absolute value terms act like the harmonic series and will diverge.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\sum_{k\ =\ 0}^{\infty}{\pars{-1}^{k} \over 2k + 1}
&=\sum_{k\ =\ 0}^{\infty}
\bracks{{1 \over 2\pars{2k} + 1} - {1 \over 2\pars{2k + 1} + 1}}
=\sum_{k\ =\ 0}^{\infty}{2 \over \pars{4k + 1}\pars{4k + 3}}
\\[5mm]&={1 \over 8}\sum_{k\ =\ 0}^{\infty}{1 \over \pars{k + 1/4}\pars{k + 3/4}}
\end{align}

When $\ds{k \ggg 1}$, the series general term goes like $\ds{1 \over k^{2}}$ such that it converges as the
  Basel Problem.

Indeed,
$$
\sum_{k\ =\ 0}^{\infty}{\pars{-1}^{k} \over 2k + 1}
<{2 \over 3} + {1 \over 8}\sum_{k\ =\ 1}^{\infty}{1 \over k^{2}}
$$

It can be evaluated as:
  $$
\color{#66f}{\large\sum_{k\ =\ 0}^{\infty}{\pars{-1}^{k} \over 2k + 1}}
={1 \over 8}\,{\Psi\pars{3/4} - \Psi\pars{1/4} \over 3/4 - 1/4}
={1 \over 4}\bracks{\pi\cot\pars{\pi\,{1 \over 4}}} =
\color{#66f}{\large{\pi \over 4}}\approx{\tt 0.7854}
$$

A: you can use the Alternating Series Test,
http://en.wikipedia.org/wiki/Alternating_series_test
