Need to clarify that this series diverges

Question

$$\sum_{n=3}^\infty \frac{1}{\log^2 n}$$ diverges by using integral test.

Is this true?

Here is the main result for integral test. – Mhenni Benghorbal Jun 11 '13 at 05:07 — Mhenni Benghorbal, Jun 11 '13 at 05:07

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

Referring to the main result, the integral test goes like this, pick up the function $f(x)= \frac{1}{\ln(x)^2} $ and see this

$$ \int_{3}^{\infty}\frac{dx}{\ln(x)^2} = \int_{\ln(3)}^{\infty}\frac{e^y}{y^2}dy, $$

where the last integral is a divegent integral, since $\lim_{y\to \infty}\frac{e^y}{y^2}=\infty$.

Note: We used the change of variables $y=\ln(x)$ in the above integral.

edited Apr 13 '17 at 12:21

Community

answered Jun 11 '13 at 05:19

Mhenni Benghorbal

score 1 · Answer 2 · answered Jun 11 '13 at 05:07

1

For $n\ge 1$, $\ln(n)^2<n$ so this series diverges by comparison to $\sum 1/n$.

answered Jun 11 '13 at 05:07

Jonathan

I agree, but OP said "using integral test". – Did Jun 11 '13 at 19:36
Oh, I couldn't really understand that from the posed question – Jonathan Jun 11 '13 at 22:29

2 Answers2