Question

Provide an example of two series , and b that both diverge, but (a -b.) converges.

Answer 1

Let the sequence a_n be equal to:

$a_n=1+\frac{1}{n^2}$

And the sequence b_n be equal to:

$b_n=1$

Notice that both series diverge:

$\begin{gathered} \sum ^{\infty}_{n\mathop=1}a_n=\sum ^{\infty}_{n\mathop=1}(1+\frac{1}{n^2})\rightarrow\infty \\ \sum ^{\infty}_{n\mathop=1}b_n=\sum ^{\infty}_{n\mathop=1}(1)\rightarrow\infty \end{gathered}$

Nevertheless, notice that the sequence a_n-b_n is given by:

$\begin{gathered} a_n-b_n=(1+\frac{1}{n^2})-1 \\ =\frac{1}{n^2} \end{gathered}$

Then:

$\sum ^{\infty}_{n\mathop=1}(a_n-b_n)=\sum ^{\infty}_{n\mathop=1}\frac{1}{n^2}$

It is a famous result that:

$\sum ^{\infty}_{n\mathop=1}\frac{1}{n^2}=\frac{\pi^2}{6}$

Then we found two divergent series so that each series diverge and the series of the difference a_n-b_n converge. The series are:

$\begin{gathered} \sum ^{\infty}_{n\mathop=1}(1+\frac{1}{n^2}) \\ \sum ^{\infty}_{n\mathop=1}(1) \end{gathered}$