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Absolute convergence

Alexmahone

Active member
Jan 26, 2012
268
Prove that if $\displaystyle \left|\frac{a_{n+1}}{a_n}\right|\le\left|\frac{b_{n+1}}{b_n}\right|$ for $\displaystyle n\gg 1$, and $\displaystyle\sum b_n$ is absolutely convergent, then $\displaystyle\sum a_n$ is absolutely convergent.
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Prove that if $\displaystyle \left|\frac{a_{n+1}}{a_n}\right|\le\left|\frac{b_{n+1}}{b_n}\right|$ for $\displaystyle n\gg 1$, and $\displaystyle\sum b_n$ is absolutely convergent, then $\displaystyle\sum a_n$ is absolutely convergent.
Isn't this obvious from the comparison test?
 

Alexmahone

Active member
Jan 26, 2012
268
Isn't this obvious from the comparison test?
My attempt:

Pick a $\displaystyle c$ large enough so that $\displaystyle|a_1|<c|b_1|$.

Assume that $\displaystyle|a_n|<c|b_n|$ for some $n$.

$\displaystyle|a_{n+1}|\le|a_n|\left|\frac{b_{n+1}}{b_n}\right|=\left|\frac{a_n}{b_n}\right||b_{n+1}|<c|b_{n+1}|$

So, by induction, $\displaystyle|a_n|<c|b_n|$ for all $n$.

Since $\displaystyle\sum c|b_n|$ converges, $\displaystyle\sum|a_n|$ also converges by the comparison test.

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Does that look ok?
 
Last edited:
Jan 31, 2012
54
My attempt:

Pick a $\displaystyle c$ large enough so that $\displaystyle|a_1|<c|b_1|$.

Assume that $\displaystyle|a_n|<c|b_n|$ for some $n$.

$\displaystyle|a_{n+1}|\le|a_n|\left|\frac{b_{n+1}}{b_n}\right|=\left|\frac{a_n}{b_n}\right||b_{n+1}|<c|b_{n+1}|$

So, by induction, $\displaystyle|a_n|<c|b_n|$ for all $n$.

Since $\displaystyle\sum c|b_n|$ converges, $\displaystyle\sum|a_n|$ also converges by the comparison test.

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Does that look ok?
Yes. And if to be more accurate your $c$ is $\frac{a_k}{b_k}$ for some natural $k$.
 

Alexmahone

Active member
Jan 26, 2012
268
Jan 31, 2012
54
What do you mean?
Say that, for some n \geq k, the followinf true:

$$ \frac{a_{n+1}}{a_n}\le\frac{b_{n+1}}{b_n} $$

We'll write:

$$ \frac{a_{k+1}}{a_k}\le\frac{b_{k+1}}{b_k} $$

$$ \frac{a_{k+2}}{a_{k+1}}\le\frac{b_{k+2}}{b_{k+1}} $$

$$ \frac{a_{k+3}}{a_{k+2}}\le\frac{b_{k+3}}{b_{k+2}} $$

.
.
.

$$ \frac{a_{n}}{a_{n-1}}\le\frac{b_{n}}{b_{n-1}} $$


Now, if we multiply these inqualities, we'll get:

$$ \frac{a_n}{a_k}\le\frac{b_n}{b_k} $$

or:

$$a_n\le\frac{a_k}{b_k}b_n$$