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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.

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  • $\begingroup$ In your hint, should the indices in the sums be $k$ instead of $n$? $\endgroup$ – Antonio Vargas Mar 17 '13 at 14:23
  • $\begingroup$ What did you try? $\endgroup$ – Did Mar 17 '13 at 14:26
  • $\begingroup$ Your proof is a little hard to follow, but otherwise, it looks correct. $\endgroup$ – robjohn Mar 17 '13 at 16:51
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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.

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  • $\begingroup$ thanks. could u help me check if my above ans is correct? $\endgroup$ – Joyce Mar 17 '13 at 15:57
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    $\begingroup$ your answer is hard to unterstand, try to rephrase it $\endgroup$ – Dominic Michaelis Mar 17 '13 at 16:06
  • $\begingroup$ ok, i've tried to rephrase. so sorry about that. $\endgroup$ – Joyce Mar 17 '13 at 16:37
  • $\begingroup$ @LeongLyn, you came here to learn. I believe you learnt something, there is nothing to be sorry about. $\endgroup$ – vonbrand Mar 17 '13 at 16:39
  • $\begingroup$ @LeongLyn you should really mention somewhere, that you assume $T_{2n}$ to be convergent, and that sum of convergent series is convergent, and as vonbrand said really no reason to be sorry $\endgroup$ – Dominic Michaelis Mar 17 '13 at 16:40
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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.

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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.

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