I am given that a series follows the following formula:
$$\sum_{n=1}^{\infty} 1/\sqrt{(n^2+n)} %$$
I approxiamate it with the following integral: $$\int_1^∞ 1/x \, dx = ln(∞)-ln(1)$$ Which simplifies to infinity.
The answer says this is correct, but my question is, since $$1/\sqrt{(n^2+n)}<1/x$$ how does the integral prove anything? The fact 1/x diverges has no bearing on $$1/\sqrt{(n^2+n)}$$ since 1/x lies above it. Where is my reasoning wrong?
It can be shown that for a continuous function $f:\mathbb{[1,\infty)}\rightarrow[0,\infty)$, the convergence of $$\sum\limits_{n=1}^\infty f(n)$$ is equivalent to the convergence of $$\int\limits_1^\infty f(x) d x.$$
So to assess whether or not your series converges, you set $f(n) = \frac{1}{\sqrt{n^2+n}}$ and look at the following integral: $$\int\limits_1^\infty \frac{1}{\sqrt{x^2+x}} dx,$$
Since for all $x\geq 1$ we have $\frac{1}{2x^2} \leq \frac{1}{x^2+x}$, we can use the smaller function to get a bound from below for the integral in question:
$$ \int\limits_1^\infty \frac{1}{\sqrt{x^2+x}} dx \geq \int\limits_1^\infty \frac{1}{\sqrt{2x^2}} dx = \frac{1}{\sqrt{2}}\int\limits_1^\infty \frac{1}{x} dx = \infty$$