I see that $\sum (x_n + y_n) = \sum x_n + \sum y_n$, but I'm not sure what formal logic I should use to prove the convergence of $\sum (x_n + y_n)$ from the convergence of the other two series.

Does anyone have any pointers? Thanks!

  • $\begingroup$ Well if $\sum (x_{n}+y_{n})$ converges absolutely then the decompostion is fine, at least $\endgroup$ – TheOscillator Nov 19 '15 at 19:46
  • $\begingroup$ Hint: let $s_n=\sum_{1}^{n}x_n, e_n=\sum_{1}^{n}y_n$ $\endgroup$ – R.N Nov 19 '15 at 19:50
  • $\begingroup$ If $a_n\to a,b_n\to b$, then $a_n+b_n\to a+b$. $\endgroup$ – Yai0Phah Nov 19 '15 at 19:50
  • $\begingroup$ if using Weistrass definition of convergence that epsilon delta. what you proving is in many real anysis books or advanced calc. Im sure you can find it $\endgroup$ – Tiger Blood Nov 19 '15 at 19:53

That $\sum\limits_{n=1}^\infty {x_n}$ converges means that the sequence $\Big(\sum\limits_{n=1}^m x_n\Big)_{m=1,2,\dots}$ converges as $m\to\infty$.

Recall the following fact.

If two sequences $(a_m)_m$ and $(b_m)_m$ converge, then their sum $(a_m+b_m)_m$ also converges (to the sum of the limits).

Thus, if $\sum\limits_{n=1}^\infty a_n$ and $\sum\limits_{n=1}^\infty b_n$ converge, then also the sequence $$\Big(\sum_{n=1}^m a_n + \sum_{n=1}^m b_n\Big)_m=\Big(\sum_{n=1}^m (a_n + b_n)\Big)_m$$ converges. In other words, $\sum\limits_{n=1}^\infty (a_n+b_n)$ converges.

  • $\begingroup$ I see the general idea, but I'm just getting a little confused by the notation. You mention the sequence $(\sum_{n=1}^m a_n + \sum_{n=1}^m b_n)_m$, but I'm not really sure why we are constructing a sequence from the series or what that means. Do you mean the sequence of partial sums of the series? $\endgroup$ – mxdg Nov 19 '15 at 19:59
  • $\begingroup$ Yes ,the series $\sum_{n=1}^\infty a_n$ is defined as the limit of the sequence of partial sums, which is given by $(\sum_{n=1}^m a_n)_m$. $\endgroup$ – J.R. Nov 19 '15 at 20:05

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