I proved it as follows.

Since $\sum a_n$ and $ \sum b_n $ are divergent,

$ \forall \epsilon > 0, \exists p \in \mathbb Z_+ st, n \gt p \implies \sum a_n > \epsilon \gt \frac{\epsilon}{2} $

And $ \forall \epsilon \gt 0, \exists q \in \mathbb Z_+ st, n \gt q \implies \sum b_n \gt \epsilon \gt \frac{\epsilon}{2} $

Let $r = max \{p,q\}$

$n \gt r \implies x \gt p$ and $x \gt q \implies \sum a_n \gt \frac{\epsilon}{2} \text{ and }\sum b_n > \epsilon /2 \implies (\sum b_n ) + (\sum a_n) > \epsilon \implies \sum (a_n + b_n ) \gt \epsilon$

Is there a fault in this?

  • 3
    $\begingroup$ What if $a_n = -b_n$? $\endgroup$
    – William
    Jul 29, 2014 at 15:23
  • $\begingroup$ It seems that you assume $\sum a_n = +\infty$ and $\sum b_n = +\infty$, is that the case? Are $a_n$, $b_n$ supposed to be positive? $\endgroup$ Jul 29, 2014 at 15:28
  • $\begingroup$ Do you assume that $a_n,b_n\ge 0$? $\endgroup$ Jul 29, 2014 at 15:28
  • $\begingroup$ I have stated that $b_n > \epsilon \forall \epsilon >0 $ , Is it wrong? $\endgroup$
    – S.Dan
    Jul 29, 2014 at 15:30

2 Answers 2


If a series is divergent, this does not imply the line

$\forall \epsilon > 0$, $\exists p \in \mathbb{Z}_{+}$ s.t. $n > p \implies \sum a_n > \epsilon > \epsilon/2$

(although I'm pretty sure I know what you mean here, this is syntactically a mess; I think you mean to say $\sum_{k = 1}^n a_k > \epsilon$). This line (when corrected) would be saying that the only way a series can diverge if it's unbounded - which is false. After all, the series

$$\sum_{n = 1}^{\infty} (-1)^n$$

is clearly divergent, but all the partial sums are either $-1$ or $0$ - hence bounded.

The claim is actually false, unless one assumes $a_n, b_n \ge 0$ - in that case, your proof is just about fine.


Consider $\sum \frac 1n$ and $\sum \frac {-1}{n}$. Both are divergent, but the 'sum-series' is 0 hence convergent. (disproved)


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