Does the series $\sum_{n=1}^\infty{c_n}$ necessarily diverge? Justify. Suppose that the series $\sum\limits_{n=1}^\infty{a_n}$ is convergent and the series $\sum\limits_{n=1}^\infty{b_n}$ is divergent. Let $c_{2n-1}$ = $a_n$ and $c_{2n}$ = $b_n$. 
Does the series $\sum\limits_{n=1}^\infty{c_n}$ necessarily diverge? Justify.
Hint: Let $R_n$ = $\sum\limits_{k=1}^n{a_n}$ , $S_n$ = $\sum\limits_{k=1}^n{b_n}$ , and $T_n$ = $\sum\limits_{k=1}^n{c_n}$. Then $T_{2n}$ = $R_n$ + $S_n$. Is $T_{2n}$ convergent?
How to link the hint to the question?
Please help me check if my proof is correct:
Since $\sum\limits_{n=1}^\infty{a_n}$ is convergent, $R_n$ = $\sum\limits_{k=1}^n{a_n}$ is also convergent.
Since $\sum\limits_{n=1}^\infty{b_n}$ is divergent, $S_n$ = $\sum\limits_{k=1}^n{b_n}$ is also divergent.
Proof (by contradiction):
Assume that $T_{2n}$ = $R_{n}$ + $S_{n}$ (by hint) is convergent.
Since $\sum\limits_{k=1}^n(-1){a_n}$ = -$\sum\limits_{k=1}^n{a_n}$  converges,
  -$\sum\limits_{k=1}^n{a_n}$ + $\sum\limits_{k=1}^n{a_n+b_n}$ = $\sum\limits_{k=1}^n{b_n}$
This implies that $\sum\limits_{k=1}^n{b_n}$ converges.
Thus, there exists a contradiction since $\sum\limits_{k=1}^n{b_n}$ diverges.
This implies that $T_{2n}$ is divergent.
Therefore, $T_{n}$ diverges, which implies that $\sum\limits_{k=1}^\infty{c_n}$ diverges.
 A: Lets define 
$$R_n= \sum_{k=0}^n a_n \qquad S_n=\sum_{k=0}^n b_n \qquad T_n = \sum_{k=0}^n c_n$$
As
$$T_{2n}=R_n + S_n$$ 
we know that 
$$T_{2n}-R_n=S_n$$
If $T_{2n}$ is convergent, we would have convergent minus convergent equals divergent, which is impossible, hence $T_{2n}$ is divergent and hence $T_n$ is divergent. 
A: Hint: consider the series
$$
\sum_{n=1}^\infty(c_{2n-1}+c_{2n})
$$
More:
$$
\begin{align}
T_{2n}
&=\sum_{k=1}^n(c_{2k-1}+c_{2k})\\
&=\sum_{k=1}^n(a_k+b_k)\\
&=R_n+S_n
\end{align}
$$
Assume that $T_n$ converges; then $T_{2n}$ also converges (any subsequence of a convergent sequence converges). Thus, $S_n=T_{2n}-R_n$ is convergent (the difference of two convergent sequences is convergent). However, $S_n$ does not converge, therefore, our assumption is false. That is, $T_n$ is divergent.
A: Can $T_{2n}$ be convergent? 
Assume it is so. Then, since $R_n$ is also convergent we have $T_{2n}-R_n = S_n$ is also convergent. Which means $\sum\limits_{n=1}^\infty{b_n}$ is convergent. This is a contradiction. 
Therefore $T_{2n}$ is divergent. Hence, $T_{n}$ is divergent. Which means $\sum\limits_{n=1}^\infty{c_n}$ is divergent. 
