When to use the limit comparison test

Question

$$\int_{7}^{\infty}\frac{1}{\sqrt{v-5}}$$ I tried using the limit comparison test and found that $\frac{1}{\sqrt{v}}\leq\frac{1}{\sqrt{v-5}}$ and therefore tried t find $\lim_{x\to\infty}\frac{\frac{1}{\sqrt{v}}}{\frac{1}{\sqrt{v-5}}}$ and found that it converges as the limit is 1 . However, the answer is diverge . Am i doing it right ?

Should i have used the Direct Comparison Test instead or could the answer also be solved by using the limit comparison test? because i'm not quite sure when to use which one

Answer 1

You were almost there. Indeed, you forgot the last step:

$$\int_7^\infty\frac1{\sqrt v}dv\to\infty$$

Thus, it diverges.

Answer 2

Your first paragraph mixes two tests: the direct comparison and the limit comparison. You can do this problem either way (in fact, you can just attack the problem directly, without either test), but what you've written makes little sense. Indeed you seem to be confusing the issue of the convergence of the integral with the issue of the convergence of the limit involved in the limit comparison test.

If you want to use the direct comparison test, just use the inequality you noticed: $1/\sqrt{v}\leq 1/\sqrt{v-5}$. The direct comparison test then says that if the integral of $1/\sqrt{v}$ diverges, so does your integral. And indeed the integral of $1/\sqrt{v}$ does diverge (this can be checked directly). Therefore your original diverges.

The limit comparison test, by contrast, says that if the limit you calculated is some positive real number, then both integrals converge or both diverge. Since the limit you calculated is 1, which is positive, the hypothesis of the test is satisfied, and the correct conclusion is that your two integrals either both converge or both diverge. To finish, you then need to distinguish between these two cases. Since we already know the integral of $1/\sqrt{v}$ diverges, we conclude your original integral does, too. But this is overkill since a direct comparison readily presented itself.