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

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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Debbie Moore
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
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