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

calculusconvergence-divergencediscrete mathematicsproof-verificationsequences-and-series

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.

Best Answer

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.

Related Solutions

[Math] A series such that $\sum {a_n}$ converges, but $\sum {a_{3n}}$ diverges.

It is entirely fine to define a sequence term by term, and your examples are fine. In fact $\LaTeX$ even supports this with the following environment:

a_n=
\begin{cases}
 [value 1] & [condition 1] \\
 [value 2] & [condition 2] \\
 ...
\end{cases}

For example (right click to show underlying code): $$a_n= \begin{cases} 2&\text{ if }\ 3\text{ divides } n\\ -1&\text{ otherwise} \end{cases} \qquad\text{ and }\qquad b_n= \begin{cases} 0&\text{ if }\ 3\text{ divides } n\\ 1&\text{ otherwise} \end{cases}.$$

Let $a_n>0$; $\sum a_n$ diverges; find $b_n$ s.t. $b_n>0$; $b_n/a_n\to0$; $\sum b_n$ diverges

Note that the sequence of partial sums $S_n = \sum_{k=1}^n a_k$ is increasing and divergent. Hence,

$$\left|\sum_{k=n+1}^m b_k \right|= \sum_{k=n+1}^m \frac{a_k}{S_k} > \frac{1}{S_m}\sum_{k=n+1}^m a_k = \frac{S_m- S_n}{S_m} = 1 - \frac{S_n}{S_m}$$

Since $S_m \to \infty$ as $m \to \infty$, for any fixed $n$ there exists $m > n$ such that $S_n/S_m < 1/2$ and

$$\left|\sum_{k=n+1}^m b_k \right| > \frac{1}{2}$$

Therfore, $\sum b_n$ diverges as the Cauchy criterion is violated.

Best Answer

Related Solutions

[Math] A series such that $\sum {a_n}$ converges, but $\sum {a_{3n}}$ diverges.

Let $a_n>0$; $\sum a_n$ diverges; find $b_n$ s.t. $b_n>0$; $b_n/a_n\to0$; $\sum b_n$ diverges

Related Question