Analytic Proof of alternating series Having some problems figuring out the following proof. If we have a series $a_n$ that is both nonnegative and decreasing. Then if we consider the corresponding alternating series $$\sum (-1)^{n+1}a_n$$ I need to prove that the sequence of odd partial sums $$s_{2n+1}= \sum _{k=1}^{2n+1} (-1)^{k+1}a_k $$ is bounded and decreasing.
So if we start to write out some terms of this series we see the following,
$(-1)^2a_1+(-1)^3a_2+(-1)^5a_4+...+(-1)^{2n+2}a_k$
So the terms seem to always equal 1 but how do I show that this is bounded and decreasing?
 A: Let
$A_n
=\sum_{k=1}^n (-1)^{k+1}a_k
$.
Then
$\begin{array}\\
A_{n+2}
&=\sum_{k=1}^{n+2} (-1)^{k+1}a_k\\
&=\sum_{k=1}^{n} (-1)^{k+1}a_k+(-1)^{n+2}a_{n+1}+(-1)^{n+3}a_{n+2}\\
&=A_n+(-1)^{n+2}(a_{n+1}-a_{n+2})\\
\end{array}
$
Putting $2n$
for $n$,
$\begin{array}\\
A_{2n+2}
&=A_{2n}+(-1)^{2n+2}(a_{2n+1}-a_{2n+2})\\
&=A_{2n}+(a_{2n+1}-a_{2n+2})\\
&>A_{2n}
\qquad\text{since }a_{2n+1}>a_{2n+2}\\
\end{array}
$
Putting $2n+1$
for $n$,
$\begin{array}\\
A_{2n+3}
&=A_{2n+1}+(-1)^{2n+3}(a_{2n+2}-a_{2n+3})\\
&=A_{2n+1}-(a_{2n+2}-a_{2n+3})\\
&<A_{2n+1}
\qquad\text{since }a_{2n+2}>a_{2n+3}\\
\end{array}
$
Therefore
the even terms are increasing
and
the odd terms are decreasing.
To show that
the odd terms are bounded below,
$\begin{array}\\
A_{2n+1}
&=\sum_{k=1}^{2n+1} (-1)^{k+1}a_k\\
&=\sum_{k=1}^{2n} (-1)^{k+1}a_k+(-1)^{2n+2}a_{2n+1}\\
&=\sum_{k=1}^{n} ((-1)^{2k}a_{2k-1}+(-1)^{2k+1}a_{2k})+a_{2n+1}\\
&=\sum_{k=1}^{n} (-1)^{2k}(a_{2k-1}-a_{2k})+a_{2n+1}\\
&=\sum_{k=1}^{n} (a_{2k-1}-a_{2k})+a_{2n+1}\\
&> 0\\
\end{array}
$
since all terms are positive.
To show that
the even terms are bounded above,
$\begin{array}\\
A_{2n+1}-A_{2n}
&=(-1)^{2n+2}a_{2n+1}\\
&=a_{2n+1}\\
&> 0\\
\end{array}
$
so
$A_{2n}
< A_{2n+1}
$.
Since the odd terms
are bounded below and decreasing,
the even terms are
bounded above.
