Comparison of the odd terms of two convergent series Let $\sum \alpha_k$ and $\sum \beta_k$ be two convergent series with strictly positive and strictly decreasing terms, i.e. $1=\alpha_1>\alpha_2>\alpha_3> \cdots > 0$ and $1=\beta_1>\beta_2>\beta_3> \cdots > 0$. 
I am trying to prove that $\sum_{k=1}^\infty \alpha_k < \sum_{k=1}^\infty \beta_k$ implies $\sum_{k=0}^\infty \alpha_{2k+1} < \sum_{k=0}^\infty \beta_{2k+1}$; however I could only prove that $\sum_{k=0}^\infty \alpha_{2k+1} < \frac{1}{2} + \sum_{k=0}^\infty \beta_{2k+1}$ by starting with $2\times\sum_{k=0}^\infty \alpha_{2k+1} $ and using the inequalities hypothesis in the reasoning. I am wondering if the statement is true or there is a counterexample ?
 A: Select $a_n$ so that $\sum a_k$ converges and is decreasing.  Let $b_n = a_n$ if $n \ne 2$.  Let $a_1 > b_2 > a_2$.
The $\sum a_k = \sum b_k -(b_2 - a_2)$ but $\sum a_{2k+1} = \sum b_{2k+1}$
A: There's an easy counterexample for the following statement: 

Let $\sum \alpha_k$ and $\sum \beta_k$ be two convergent series with strictly positive and nonincreasing terms, i.e. $1=\alpha_1 \ge \alpha_2 \ge \alpha_3 \ge \cdots > 0$ and $1=\beta_1 \ge \beta_2 \ge \beta_3 \ge \cdots > 0$. Then  $$ \sum_{k=1}^\infty \alpha_k \le \sum_{k=1}^\infty \beta_k 
\Longrightarrow
\sum_{k=0}^\infty \alpha_{2k+1} \le \sum_{k=0}^\infty \beta_{2k+1}. $$

Choose the following sequences 
$$ \alpha : \quad 1,2^{-2},2^{-2},2^{-3},2^{-3},\ldots, $$
$$ \beta : \quad 1,2^{-1},2^{-2},2^{-3},2^{-4},\ldots. $$
Then 
$$ \sum_{k=1}^\infty \alpha_k = 2 = \sum_{k=1}^\infty \beta_k, $$
$$ \sum_{k=0}^\infty \alpha_{2k+1} = \frac 32 > \frac 43 = \sum_{k=0}^\infty \beta_{2k+1}. $$
Now we just need to alter $\alpha_k$ a little. Choose $0<\varepsilon<\frac 13$, then let 
$$ \bar{\alpha}_1 = 1, \ \bar{\alpha}_{2k} = \alpha_{2k}, \ \bar{\alpha}_{2k+1} = (1-\varepsilon) \alpha_{2k+1} \text{ for } k \ge 1. $$
We have
$$ \sum_{k=1}^\infty \bar{\alpha}_k = 2 - \frac{\varepsilon}{2} < 2 =  \sum_{k=1}^\infty \beta_k $$
and 
$$ \sum_{k=0}^\infty \bar{\alpha}_{2k+1} = \frac 32 - \frac{\varepsilon}{2} > \frac 43 = \sum_{k=0}^\infty \beta_{2k+1}. $$
due to $\varepsilon < \frac 13$. Since $0 < \varepsilon < \frac 12$, this sequence is also strictly decreasing. 
