Week 3 - Convergence Tests

• Ratio Test
  • Example
  • Theorem
  • Other Examples
• Root Test
  • Theorem
• Integral Test
  • Theorem
  • Proof of the Integral Test
  • Example
• Words

1. Ratio Test

1.1. Example

• Does $\sum\frac{n^5}{4^n}$ convergence?
  • Use the Geometric Series Theory, we know that $\sum\frac{1}{4^n}$ convergence, because $\sum\frac{1}{4^n} = \sum\frac{1}{4} \cdot {\frac{1}{4}^{n-1} }$ and $\frac{1}{4} < 1$.
  • Last week, we've learned Comparison Test Theory. So if we can find a series that every element in it is bigger than this one and also converges, then this example must converge, too.
  • Let's calculate $\lim_{n \to \infty}\frac{a_n}{a_{n-1} }$
    • $\lim_{n \to \infty}\frac{a_n}{a_{n-1} } = \lim_{n \to \infty} \frac{\frac{n^5}{4^n} }{\frac{(n-1)^5}{4^{n-1} } } = \lim_{n \to \infty}\frac{n^5}{4^n} \cdot \frac{ 4^{ n-1 } }{ (n-1)^5 } = \lim_{n \to \infty}\frac{1}{4} (\frac{n}{n-1})^5 = \frac{1}{4}$
    • So if n is big enough, then the common ratio will be close to 1/4.
  • Then we can pick a constant $\epsilon = 0.1$ that make $\vert \frac{a_n}{a_{n-1} } \vert < \frac{1}{4} + 0.1$, after some calculation we got that $n \ge 15$.
    • So, $a_{n+1} < (\frac{1}{4} + 0.1) a_n,\ \text{where}\ n \ge 15$,
    • Then $a_{15+k} < (\frac{1}{4} + 0.1)^k a_{15}$.
  • We know that $\sum (\frac{1}{4} + 0.1)^k a_{15}$ convergence, base on the Geometric Series Theory.
  • So $\sum a_{15+k}$ converge.
  • So $\sum a_{k}$ converge.

1.2. Theorem

• Consider $\displaystyle \sum_{n=0}^{\infty} a_n$. And $a_n \ge 0$, $\displaystyle \lim_{n \to \infty} \frac{a_{n+1} }{a_n} = L$.
  • If $L < 1$, then $\displaystyle \sum_{n=0}^{\infty} a_n$ converges.
  • If $L > 1$, then $\displaystyle \sum_{n=0}^{\infty} a_n$ diverges.
  • If $L = 1$, then the ratio test is inconclusive.

Detailed

• To find the series ($\sum_{n=1}^{\infty}a_n,\ \text{where}\ a_n > 0$) converge or not.
• If $\lim_{n \to \infty}\frac{a_{n+1} }{a_n} = L < 1$,
  • Then pick a small $\epsilon$ so that $L + \epsilon < 1$
  • Find N so that $n \ge N$, $\frac{a_{n+1} }{a_n} < L + \epsilon$.
  • Then $a_{N+k} < (L+\epsilon)^k \cdot a_{N}$
  • Since $\sum (L+\epsilon)^k \cdot a_{N}$ converge, $\sum a_{N+k}$ converge.
  • So $\sum a_{k}$ converge.
• If $\lim_{n \to \infty}\frac{a_{n+1} }{a_n} = L > 1$,
  • Then pick a small $\epsilon$ so that $L - \epsilon > 1$
  • Find N so that $n \ge N$, $\frac{a_{n+1} }{a_n} > L - \epsilon$.
  • Then $a_{N+k} > (L-\epsilon)^k \cdot a_{N}$
  • Since $\sum (L+\epsilon)^k \cdot a_{N}$ diverge, $\sum a_{N+k}$ diverge.
  • So $\sum a_{k}$ diverge.
• If $\lim_{n \to \infty}\frac{a_{n+1} }{a_n} = L = 1$,
  • Might converge, like $\sum \frac{1}{n^2}$
  • Might diverge, like $\sum 1$

1.3. Other Examples

• $\sum \frac{n!}{n^n}$, $\sum \frac{n!}{(n/2)^n}$, $\sum \frac{n!}{(n/3)^n}$
• We know that $e = \displaystyle\lim_{n \to \infty} (1+\frac{1}{n})^n$. With this, we can conclude $\sum \frac{n!}{(n/2)^n}$ converges, $\sum \frac{n!}{(n/3)^n}$ diverges.
• But what about $\sum \frac{n!}{(n/e)^n}$. Let's compare $n!$ with $n^n$
  • $\log(n!) = \sum_{k=1}^n \log(k) \approx \int_1^n \log x dx = x \log x - x ]_1^n$
  • $\log(n!) = n \log n - n + 1$
  • $\log(n!) \approx n \log n - n = \log(\frac{n^n}{e^n})$
  • $n! \approx (\frac{n}{e})^n$
  • Better approximation is $n! \approx (\frac{n}{e})^n \sqrt{2 \pi n}$, which name is Stirling's approximation.

2. Root Test

2.1. Theorem

• $a_n > 0, L = \displaystyle \lim_{n \to \infty} \sqrt[n]{a_n}$
  • If $L < 1, \displaystyle \sum_{n=1}^{\infty}$ converge,
  • If $L > 1, \displaystyle \sum_{n=1}^{\infty}$ diverge,
  • If $L = 1, \displaystyle \sum_{n=1}^{\infty}$ inconclusive.

3. Integral Test

3.1. Theorem

• Suppose f is a continuous, positive, decreasing function on $[1, \infty)$, and let $a_n = f(n)$. Then the series $\sum_{n=1}^{\infty} a_n$ is convergent if and only if the improper integral $\int_{1}^{\infty} f(x) dx$ is convergent. In other words:
  • If $\int_{1}^{\infty} f(x) dx$ is convergent, then $\sum_{n=1}^{\infty} a_n$ is convergent.
  • If $\int_{1}^{\infty} f(x) dx$ is divergent, then $\sum_{n=1}^{\infty} a_n$ is divergent.

3.2. Proof of the Integral Test

• 
• For the general series $\sum a_n$, look at the figures above. The area of the first shaded rectangle is the value of f at the right endpoint of [1, 2], that is, $f(2) = a_2$. So, comparing the areas of the shaded rectangles with the area under $y = f(x)$ from 1 to n, we see that
  • $\displaystyle a_2 + \ldots + a_n = s_n - a_1 = \sum_{i=2}^n a_i \le \int_1^n f(x) dx \le a_1 + \ldots + a_{n-1} = s_{n-1} \le \sum_{i=1}^n a_i$
    • Notice that this inequality depends on the fact that f is decreasing.
• If $\int_1^n f(x) dx$ is convergent, then $\sum_{i=2}^n a_i \le \int_1^n f(x) dx \le \int_1^{\infty} f(x) dx$ since $f(x) \ge 0$. Therefore $s_n = a_1 + \sum_{i=2}^n a_i \le a_1 + \int_1^{\infty} f(x) dx$
  • So the sequence $\{s_n\}$ is bounded above.
  • And $\{s_n\}$ is also an increasing sequence.
  • This means that $\sum a_n$ is convergence.
• If $\int_1^n f(x) dx$ is divergent, then $\int_1^n f(x) dx \to \infty \ \text{as}\ n \to \infty$ because $f(x) \ge 0$. But $\displaystyle\int_1^n f(x) dx \le \sum_{i=1}^{n-1}a_i = s_{n-1}$ and so $s_{n-1} \to \infty$. This implies that $s_n \to \infty$ and $\sum a_n$ diverges.

3.3. Example

• The Harmonic Series $\sum \frac{1}{n}$
  • The Integral Test need the function is positive and decreasing on $[1, \infty)$.
    • $x \ge 1, f(x) > 0$
    • $a > b, f(a) < f(b)$
  • Now let's calculate $\int_1^{\infty} f(x) d(x)$
    • $= \displaystyle \lim_{N \to \infty} \int_1^N f(x) dx = \lim_{N \to \infty} \int_1^N \frac{1}{x} dx$
    • $= \displaystyle\lim_{N \to \infty} (\log N - \log 1) = \lim_{N \to \infty} \log N$
    • $= \infty$
• $\sum \frac{1}{n^p}$
  • Method 1:
    • With Cauchy Condensation, we know that, $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if and only if $\sum_{n=0}^{\infty} 2^n \frac{1}{(2^n)^p}$ converges.
    • $\sum_{n=0}^{\infty} 2^n \frac{1}{(2^n)^p} = \sum_{n=0}^{\infty} \frac{1}{(2^n)^{p-1} } = \sum_{n=0}^{\infty} \frac{1}{(2^{p-1})^n }$
    • Use the Geometric Series Theory, we got the ratio is $\frac{1}{2^{p-1}}$, so if $p < 1$, it diverges; if $p > 1$, it converges.
  • Method 2, use The Integral Test:
    • Let's assume $p \ne 1$, which is The Harmonic Series we have proved before.
    • Also the function is positive and decreasing on $[1, \infty)$, so we can use the Integral Test on this.
    • $\displaystyle \int_1^{\infty} f(x) d(x) = \lim_{N \to \infty} \int_1^N \frac{1}{x^p} d(x) = \lim_{N \to \infty} (\frac{1}{-p+1} N^{-p+1} - \frac{1}{-p+1})$
    • $\displaystyle = \lim_{N \to \infty} \frac{1}{-p+1} (N^{-p+1} - 1)$
    • so if $p < 1$, it diverges; if $p > 1$, it converges.

4. Words

• factorial ['fæktɔ:riəl] adj. 因子的，阶乘的 n. [数] 阶乘