# series converges by comparison test

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 $$a ($$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.

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

$$\sum_{i=1}^\infty\frac bb=\sum_{i=1}^\infty1$$ obviously does not converge, so the comparison test cannot be used in this way.
• Why does $sum_{i=1}^\infty1$ not converge? May 29, 2019 at 8:13
• @user11015000 Because the terms don't converge to $0$. May 29, 2019 at 8:14
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 $$\alpha>1$$.