4
$\begingroup$

Prove or disprove that if $\sum a_n$ and $\sum b_n$ are both divergent, then $\sum (a_n \pm b_n)$ necessarily diverges.

I decided to disprove this by offering the example:

Let $a_n = n$.

Let $b_n = -n$.

Then $\sum (a_n + b_n) = \sum 0 = 0$, which converges.

However, my question says $\pm$, and in the $-$ case, we have $\sum (a_n - b_n) = \sum 2n = \infty$, which diverges. So, does this still count as a valid counter example or not? I partly proved it wrong so I think it's still valid.

$\endgroup$

4 Answers 4

4
$\begingroup$

If $\sum(a_n+b_n)$ and $\sum(a_n-b_n)$ both converge, then since :

$$a_n=\frac12\left((a_n+b_n)+(a_n-b_n)\right)\quad\textrm{and}\quad b_n=\frac12\left((a_n+b_n)-(a_n-b_n)\right)$$

we see that $\sum a_n$ and $\sum b_n$ both converge.

In other words, if $\sum a_n$ diverges or $\sum b_n$ diverges, then $\sum(a_n+b_n)$ diverges or $\sum(a_n-b_n)$ diverges.

$\endgroup$
2
  • 1
    $\begingroup$ Although this proves that if the original series don't both converge at least one of the $\pm$ series doesn't either, "does not converge" is not equivalent to "diverges". $\endgroup$
    – J.G.
    Commented Mar 20, 2017 at 22:38
  • $\begingroup$ Further to my already upvoted comment, apparently Wikipedia does define divergent as "not convergent", even though I could have sworn it meant "tends to $\pm\infty$". It seems weird to me that an oscillatory sequence, for example, would be described as "divergent". $\endgroup$
    – J.G.
    Commented Mar 20, 2017 at 23:00
2
$\begingroup$

My reading of the statement "$\sum (a_n \pm b_n)$ necessarily diverges" is that it means that both of the sums $\sum (a_n + b_n)$ and $\sum (a_n - b_n)$ diverge. You have proved that this is not necessarily true; you have a counterexample in which one of the sums converges.

The wording of the original question seems highly ambiguous to me, however. The statement "$\sum (a_n \pm b_n)$ diverges" might be intended to say that either $\sum (a_n + b_n)$ diverges or $\sum (a_n - b_n)$ diverges.

It really comes down to what is the meaning of the statement "$\sum (a_n \pm b_n)$ diverges."

$\endgroup$
0
$\begingroup$

Let $\sum \frac{1}{2n+1}$=$1+1/3+1/5+1/7+1/9+....$ and let $-\sum \frac{1}{2n}$=$-1/2-1/4-1/6-1/8...$. Boths series diverge, but when you add them (term by term) together, you get $1-1/2+1/3-1/4+1/5-1/6...$ and this is the well known Alternating Harmonic series with an answer of $\ln 2$, so (conditionally) convergent.

$\endgroup$
1
  • 1
    $\begingroup$ In fact, when you add them term-by-term you get $\sum \left(\frac{1}{2n-1}-\frac{1}{2n}\right)$, which is absolutely convergent as each individual term is positive. It's only when you deparenthesize that you get a conditionally convergent result... $\endgroup$
    – Micah
    Commented Mar 20, 2017 at 22:47
-1
$\begingroup$

The statement is disproved.

Given: $\sum_{n=1}^{\infty}$an and $\sum_{n=1}^{\infty}$bn are divergent.

Suppose $\sum_{n=1}^{\infty}$(an + bn) is divergent.

Let an = 1, bn = -1

Then both: $\sum_{n=1}^{\infty}$an and $\sum_{n=1}^{\infty}$bn diverge.

Now summing both series:

$\sum_{n=1}^{\infty}$(an + bn) = $\sum_{n=1}^{\infty}$(1 + (-1) = $\sum_{n=1}^{\infty}$0

which is convergent. This is a contradiction which disproves the statement.

$\endgroup$
2
  • 1
    $\begingroup$ How about $\sum (a_n-b_n)$? $\endgroup$
    – Gary
    Commented Jun 23, 2023 at 5:29
  • $\begingroup$ In the specific example above the difference diverges since 1 - (-1) = 2. So the sum converges but the difference diverges. Note that the statement states that both the sum and difference diverges. This is not true. $\endgroup$
    – syzygy
    Commented Jun 24, 2023 at 13:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .