Notes - Analysis I MT22, Integral test
Flashcards
Can you state the integral test in full?
Let $f : [1, \infty)$ be a function. Assume that
- $f$ is non-negative
- $f$ is decreasing
- $\int^{k+1} _ {k} f(x) \text{d}x$ exists for $k \ge 1$.
Let $s _ n = \sum _ {k=1}^n f(k)$ and $I _ n = \int^n _ {1}f(x)\text{d}x$. Then:
- Let $\sigma _ n = s _ n - I _ n$. Then $(\sigma _ n)$ converges, and $0 \le \lim{n\to\infty} \sigma _ n \le f(1)$.
- $(s _ n)$ converges if and only if $(I _ n)$ converges.
When proving the integral test,
- $f$ is non-negative
- $f$ is decreasing
- $\int^{k+1} _ {k} f(x) \text{d}x$ exists for $k \ge 1$.
Let $s _ n = \sum _ {k=1}^n f(k)$ and $I _ n = \int^n _ {1}f(x)\text{d}x$. Then:
- Let $\sigma _ n = s _ n - I _ n$. Then $(\sigma _ n)$ converges, and $0 \le \lim _ {n\to\infty} \sigma _ n \le f(1)$.
- $(s _ n)$ converges if and only if $(I _ n)$ converges.
what’s the strategy for showing $(\sigma _ n)$ converges?
- $f$ is non-negative
- $f$ is decreasing
- $\int^{k+1} _ {k} f(x) \text{d}x$ exists for $k \ge 1$.
Let $s _ n = \sum _ {k=1}^n f(k)$ and $I _ n = \int^n _ {1}f(x)\text{d}x$. Then:
- Let $\sigma _ n = s _ n - I _ n$. Then $(\sigma _ n)$ converges, and $0 \le \lim _ {n\to\infty} \sigma _ n \le f(1)$.
- $(s _ n)$ converges if and only if $(I _ n)$ converges.
Show it is decreasing and bounded below.
What’s a quick proof that if
\[f(k+1) \le f(x) \le f(k)\]
then
\[f(k+1) \le \int^{k+1}_k f(x) \text d x \le f(k)\]
?
\[\begin{aligned}
f(k+1) &= \int_k^{k+1} f(k+1) \text dx \\\\
&\le \int_k^{k+1} f(x) \text dx \\\\
&\le \int_k^{k+1} f(k) \text dx \\\\
&= f(k)
\end{aligned}\]
Proofs
The integral test states that when
- $f$ is non-negative
- $f$ is decreasing
- $\int^{k+1} _ {k} f(x) \text{d}x$ exists for $k \ge 1$.
Let $s _ n = \sum _ {k=1}^n f(k)$ and $I _ n = \int^n _ {1}f(x)\text{d}x$. Then:
- Let $\sigma _ n = s _ n - I _ n$. Then $(\sigma _ n)$ converges, and $0 \le \lim _ {n\to\infty} \sigma _ n \le f(1)$.
- $(s _ n)$ converges if and only if $(I _ n)$ converges.
Prove (1).
- $f$ is non-negative
- $f$ is decreasing
- $\int^{k+1} _ {k} f(x) \text{d}x$ exists for $k \ge 1$.
Let $s _ n = \sum _ {k=1}^n f(k)$ and $I _ n = \int^n _ {1}f(x)\text{d}x$. Then:
- Let $\sigma _ n = s _ n - I _ n$. Then $(\sigma _ n)$ converges, and $0 \le \lim _ {n\to\infty} \sigma _ n \le f(1)$.
- $(s _ n)$ converges if and only if $(I _ n)$ converges.
Todo.
The integral test states that when
- $f$ is non-negative
- $f$ is decreasing
- $\int^{k+1} _ {k} f(x) \text{d}x$ exists for $k \ge 1$.
Let $s _ n = \sum _ {k=1}^n f(k)$ and $I _ n = \int^n _ {1}f(x)\text{d}x$. Then:
- Let $\sigma _ n = s _ n - I _ n$. Then $(\sigma _ n)$ converges, and $0 \le \lim _ {n\to\infty} \sigma _ n \le f(1)$.
- $(s _ n)$ converges if and only if $(I _ n)$ converges.
Prove (2) using (1).
- $f$ is non-negative
- $f$ is decreasing
- $\int^{k+1} _ {k} f(x) \text{d}x$ exists for $k \ge 1$.
Let $s _ n = \sum _ {k=1}^n f(k)$ and $I _ n = \int^n _ {1}f(x)\text{d}x$. Then:
- Let $\sigma _ n = s _ n - I _ n$. Then $(\sigma _ n)$ converges, and $0 \le \lim _ {n\to\infty} \sigma _ n \le f(1)$.
- $(s _ n)$ converges if and only if $(I _ n)$ converges.
Todo.