Convergence proof: If sum to infinity of a function is convergent, does that mean for another smaller function, it will also be convergent?

Question

The logic goes something like this

Say I am able to prove that $$\int_{1}^{\infty} \frac{1}{x^{2}} dx$$ is convergent.

Now I have another expression: $$\int_{1}^{\infty} \frac{1}{x^{2}+x+1} dx $$.

For x > 0, $\frac{1}{x^{2}+x+1}$ is smaller than $\frac{1}{x^{2}}$ since the denominator is larger. Since I'm summing up smaller numbers, it follows that

$$\int_{1}^{\infty} \frac{1}{x^{2}+x+1} dx < \int_{1}^{\infty} \frac{1}{x^{2}} dx$$

Now, $\int_{1}^{\infty} \frac{1}{x^{2}+x+1} dx $ can be one of two things: it can be convergent or divergent. If it is divergent, then its sum over infinity must eventually exceed $\int_{1}^{\infty} \frac{1}{x^{2}} dx$. But we've just reasoned that it cannot happen! Thus we have a contradiction and hence $\int_{1}^{\infty} \frac{1}{x^{2}+x+1} dx $ cannot be divergent.

Hence it$\int_{1}^{\infty} \frac{1}{x^{2}+x+1} dx $ must be convergent.

Does this proof work? And does that mean that if we can prove that if $\int_{1}^{\infty} f(x)dx $ is convergent, any other function g(x) < f(x), $\int_{1}^{\infty} g(x)dx $ is convergent too? Is there a name for this proof?

Answer 1

When something is below a threshold, it doesn't mean it's convergent (in general). You also have to show that $\frac{1}{x^2+x+1}$ is eventually decreasing or increasing (but still strictly above $0$ in this case). Thereafter, you can conclude convergence. The Monotone Convergence Theorem is what you need.

Answer 2

What you have here follows from the Integral Test and the Comparison Test:

Integral Test. Let $f$ be a monotone decreasing non-negative continuous function defined on $[1, \infty)$. Then the infinite series $\sum_{n=1}^{\infty}f(n)$ converges if and only if $\int_{1}^{\infty}f(x)\text{ d}x$ is finite.

Observe that $f_1(x) = \dfrac{1}{x^2}$ is monotone decreasing and has a finite integral $\int_{1}^{\infty}f_1(x) \text{ d}x$, so that $\sum_{n=1}^{\infty}\dfrac{1}{n^2}$ converges.

Comparison Test. If the infinite series $\sum b_n$ converges and $0 \leq a_n \leq b_n$ for sufficiently large $n$, then the infinite series $\sum a_n$ converges as well.

Since $0 < \dfrac{1}{n^2 + n + 1} < \dfrac{1}{n^2}$ for all $n \geq 1$, it follows that $\sum_{n=1}^{\infty}\dfrac{1}{n^2+n+1}$ must converge.

Thus, by the integral test, since $f_2(x) = \dfrac{1}{x^2 + x + 1}$ is monotone decreasing and we have that $\sum_{n=1}^{\infty}\dfrac{1}{n^2+n+1}$ converges, $\int_{1}^{\infty}\dfrac{1}{x^2 + x + 1}\text{ d}x$ is finite.

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A more general statement.

Suppose $F$ is a monotone decreasing non-negative continuous function defined on $[1, \infty)$ and $\int_{1}^{\infty}F(x)\text{ d}x$ is finite. Let $f$ also be a monotone-decreasing non-negative continuous function defined on $[1, \infty)$ with $f(x) \leq F(x)$ for all $x$ in $[1, \infty)$. Then $\int_{1}^{\infty}f(x)\text{ d}x$ is finite.

Proof. By the integral test, $\sum_{n=1}^{\infty}F(n)$ is convergent. Since $f(n) \leq F(n)$ for all $n \geq 1$, it follows that $\sum_{n=1}^{\infty}f(n)$ is convergent. Hence, $\int_{1}^{\infty}f(x)\text{ d}x$ is convergent.