Integral Test

Theorem

• Given ${\sum }_{n=i}^{\infty }{a}_n$
• If $f(x)$ is positive
• if $f(x)$ is Continuous, on $[i,\infty )$
• If $f(x)$ is Strictly Decreasing on $[i,\infty )$
• if $\forall n\in \mathbb{N},a_n=f(n)$
• Then, ${\sum }_{n=i}^{\infty }{a}_n\text{ converges }⟺{\int }_i^{\infty }f(x)dx\text{ converges }$

Flipped Theorem

• Given ${\sum }_{n=i}^{\infty }{a}_n$
• If $f(x)$ is negative
• if $f(x)$ is Continuous, on $[i,\infty )$
• If $f(x)$ is Strictly Increasing on $[i,\infty )$
• if $\forall n\in \mathbb{N},a_n=f(n)$
• Then, ${\sum }_{n=i}^{\infty }{a}_n\text{ converges }⟺{\int }_i^{\infty }f(x)dx\text{ converges }$

Intuition

Both converge or both diverge

Examples

Note that we only need to prove $f(x)$ is strictly decreasing with derivatives as Differentiable Implies Continuity

Proof

1. Suppose $f(x)>0$ on $[1,\infty )$
2. Suppose $f(x)$ is con on $[1,\infty )$
3. Suppose $f(x)$ is Strictly Decreasing on $[1,\infty )$
4. Suppose $\forall n\in \mathbb{N},a_n=f(n)$
5. Note that $a_2(1)+\cdots +a_n(1)\le {\int }_1^{n}f(x)dx$ as it is a Right Riemann Sum, which is smaller for decreasing functions
6. Note that $a_1(1)+\cdots +a_{n-1}(1)\ge {\int }_1^{n}f(x)dx$ as it is a Left Riemann Sum, which is greater for decreasing functions

Proving $⟸$

7. Suppose ${\int }_a^{\infty }f(x)dx$ converges
8. WTS ${\sum }_{n=1}^{\infty }{a}_n$ converges.
9. Note that $a_2(1)+\cdots +a_n(1)\le {\int }_1^{n}f(x)dx$
10. $⟹a_1+a_2(1)+\cdots +a_n(1)\le a_1+{\int }_1^{n}f(x)dx$
11. $⟹S_n\le a_1+{\int }_1^{n}f(x)dx$ by definition of Series Partial Sum
12. Note that both sides are $>0$
13. It also follows that $S_n\le a_1+{\int }_1^{n}f(x)dx<a_1+{\int }_1^{\infty }f(x)dx$
14. We know that ${\int }_a^{\infty }f(x)dx$ converges, thus it is finite. Additionally, $a_1$ is a finite value. Thus, we can denote $a_1+{\int }_1^{\infty }f(x)dx=M\in \mathbb{R}$
15. Thus, $\forall n\in \mathbb{N},S_n\le M$
16. So, $S_n$ is bounded above
17. Now, we show that $S_n$ is increasing.
18. Consider $S_{n+1}=S_n+a_{n+1}$. Note that $a_{n+1}>0$ and $S_n>0$ by 1, 4
19. Thus, $S_{n+1}>S_n\forall n\in \mathbb{N}$
    1. Then, by BMCT, as $S_n$ is increasing and bounded above, our series $S_n$ converges