Suppose summation a_{n} and summation b_{n} are series with positive

Marla Payton 2021-12-19 Answered
Suppose summation \(\displaystyle{a}_{{{n}}}\) and summation \(\displaystyle{b}_{{{n}}}\) are series with positive terms and summation \(\displaystyle{b}_{{{n}}}\) is known to be divergent. If \(\displaystyle{a}_{{{n}}}{ < }{b}_{{{n}}}\) for all n, what can you say about summation \(\displaystyle{a}_{{{n}}}\)? Why?

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Expert Answer

Debbie Moore
Answered 2021-12-20 Author has 5130 answers
Step 1
There is nothing that we can say about the series \(\displaystyle\sum{a}_{{{n}}}\) It can both converge and diverge as illustrated by the next example:
Let \(\displaystyle{b}_{{{n}}}={\frac{{{1}}}{{{n}}}}\). We know that \(\displaystyle\sum{b}_{{{n}}}\) is a divergent series. First, let \(\displaystyle{a}_{{{n}}}={\frac{{{1}}}{{{n}^{{{2}}}}}}\). Now, we have \(\displaystyle{a}_{{{n}}}{ < }{b}_{{{n}}}\) for all \(\displaystyle{n}\geq{2}\), and we know that \(\displaystyle\sum{a}_{{{n}}}\) converges.
Next, let \(\displaystyle{a}_{{{n}}}={\frac{{{1}}}{{{n}+{1}}}}\) Again, we know that \(\displaystyle{a}_{{{n}}}{ < }{b}_{{{n}}}\) for all n. Using the Limit Comparison Test, we can show that \(\displaystyle\sum{a}_{{{n}}}\) diverges:
\(\displaystyle\lim_{{{n}\Rightarrow\infty}}{\frac{{{a}_{{{n}}}}}{{{b}_{{{n}}}}}}=\lim_{{{n}\Rightarrow\infty}}{\frac{{{\frac{{{1}}}{{{n}+{1}}}}}}{{{\frac{{{1}}}{{{n}}}}}}}=\lim_{{{n}\Rightarrow\infty}}{\frac{{{n}}}{{{n}+{1}}}}=\lim_{{{n}\Rightarrow\infty}}{\frac{{{\frac{{{n}}}{{{n}}}}}}{{{\frac{{{n}}}{{{n}}}}+{\frac{{{1}}}{{{n}}}}}}}\)
\(\displaystyle=\lim_{{{n}\Rightarrow\infty}}{\frac{{{1}}}{{{1}+{\frac{{{1}}}{{{n}}}}}}}={\frac{{\lim_{{{n}\Rightarrow\infty}}{1}}}{{\lim_{{{n}\Rightarrow\infty}}{1}+\lim_{{{n}\Rightarrow\infty}}{\frac{{{1}}}{{{n}}}}}}}={\frac{{{1}}}{{{1}-{0}}}}={1}\)
From here, we can conclude that \(\displaystyle\sum{a}_{{{n}}}\) is a divergent sum
Therefore, we have shown that \(\displaystyle\sum{a}_{{{n}}}\) can be both a convergent and a divergent series - thus, we cannot say anything about it in this case.
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Buck Henry
Answered 2021-12-21 Author has 4502 answers
Step 1
We cannot conclude convergence/devirgence of \(\displaystyle\sum{a}_{{{n}}}\) from the given information
Step 2
Example where \(\displaystyle\sum{a}_{{{n}}}\) converges
Let \(\displaystyle{a}_{{{n}}}={\frac{{{1}}}{{{n}^{{{2}}}}}}\) and \(\displaystyle{b}_{{{n}}}={\frac{{{1}}}{{{n}}}}\)
Both \(\displaystyle{a}_{{{n}}}\) and \(\displaystyle{b}_{{{n}}}\) are positive for \(\displaystyle{n}\geq{1}\)
\(\displaystyle{\frac{{{1}}}{{{n}^{{{2}}}}}}{ < }{\frac{{{1}}}{{{n}}}}\) for \(\displaystyle{n}\geq{1}\)
\(\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{b}_{{{n}}}={\sum_{{{n}={1}}}^{{\infty}}}{\frac{{{1}}}{{{n}}}}\) deverges
\(\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{a}_{{{n}}}={\sum_{{{n}={1}}}^{{\infty}}}{\frac{{{1}}}{{{n}^{{{2}}}}}}\) converges
Step 3
Example where \(\displaystyle{a}_{{{n}}}\) diverges
Let \(\displaystyle{a}_{{{n}}}={\frac{{{1}}}{{{2}{n}}}}\) and \(\displaystyle{b}_{{{n}}}={\frac{{{1}}}{{{n}}}}\)
Both \(\displaystyle{a}_{{{n}}}\) and \(\displaystyle{b}_{{{n}}}\) are positive for \(\displaystyle{n}\geq{1}\)
\(\displaystyle{\frac{{{1}}}{{{2}{n}}}}{ < }{\frac{{{1}}}{{{n}}}}\) for \(\displaystyle{n}\geq{1}\)
\(\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{b}_{{{n}}}={\sum_{{{n}={1}}}^{{\infty}}}{\frac{{{1}}}{{{n}}}}\) diverges
\(\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{a}_{{{n}}}={\sum_{{{n}={1}}}^{{\infty}}}{\frac{{{1}}}{{{2}{n}}}}\) diverges
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