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I am currently studying series and using the comparison test to determine whether or not a series converges.

Is this an acceptable proof using the comparison test? (I have generalised a and b)

If we know that $a<b$ ($a$ and $b$ are functions of $i$)

$$\sum_{i=1}^\infty \frac{a}{b} < \sum_{i=1}^\infty\frac{b}{b} = 1 $$

Now since a series larger than a/b converges, using the comparison test we can say that the sequence a/b converges?

Thanks in regards.

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    $\begingroup$ $\sum_{i=1}^\infty\frac{b}{b}=\sum_{i=1}^\infty1$ means to sum up infinitely many $1$ and is equal to $\infty$. $\endgroup$
    – trisct
    May 29, 2019 at 8:13
  • $\begingroup$ Oh right, that makes sense. Thank you $\endgroup$ May 29, 2019 at 8:14

2 Answers 2

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$\sum_{i=1}^\infty\frac bb=\sum_{i=1}^\infty1$ obviously does not converge, so the comparison test cannot be used in this way.

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  • $\begingroup$ Why does $sum_{i=1}^\infty1$ not converge? $\endgroup$ May 29, 2019 at 8:13
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    $\begingroup$ @user11015000 Because the terms don't converge to $0$. $\endgroup$ May 29, 2019 at 8:14
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Not only is your use of the comparison test incorrect as pointed out in the comments and answers, there is actually no way to determine if $\sum_{i=1}^\infty \frac{a_i}{b_i}$ exists.

Take $a_i = 1$ and $b_i=i^\alpha$. The series will converge if and only if $\alpha>1$.

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  • $\begingroup$ Thanks, but my use of the comparison test is fine. $\endgroup$ May 30, 2019 at 10:22

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