Respuesta :
Let the sequence a_n be equal to:
[tex]a_n=1+\frac{1}{n^2}[/tex]And the sequence b_n be equal to:
[tex]b_n=1[/tex]Notice that both series diverge:
[tex]\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}[/tex]Nevertheless, notice that the sequence a_n-b_n is given by:
[tex]\begin{gathered} a_n-b_n=(1+\frac{1}{n^2})-1 \\ =\frac{1}{n^2} \end{gathered}[/tex]Then:
[tex]\sum ^{\infty}_{n\mathop=1}(a_n-b_n)=\sum ^{\infty}_{n\mathop=1}\frac{1}{n^2}[/tex]It is a famous result that:
[tex]\sum ^{\infty}_{n\mathop=1}\frac{1}{n^2}=\frac{\pi^2}{6}[/tex]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:
[tex]\begin{gathered} \sum ^{\infty}_{n\mathop=1}(1+\frac{1}{n^2}) \\ \sum ^{\infty}_{n\mathop=1}(1) \end{gathered}[/tex]