Solution Manual of MOOCULUS-2 "Sequences and Series": 5. Another comparison test

The printable textbook of PDF format can be downloaded via MOOCULUS-2 Textbook

The printable solution manual of PDF format can be downloaded via MOOCULUS-2 Solution


Summary

• Let $N > 1$ be an integer, and consider a series $\sum_{n=1}^\infty a_n$. The series we get by removing the first $N-1$ terms, namely $$\sum_{n=N}^\infty a_n$$ is called a tail of the given series.
• Let $N > 1$ be an integer. The series $$\sum_{n=1}^\infty a_n$$ converges if and only if $$\sum_{n=N}^\infty a_n$$ converges.
  This could be shortened to "The series converges iff a tail of the series converges," or even just to the slogan that convergence depends on the tail.
• Limit Comparison Test
  Suppose $a_n \geq 0$ and $b_n \geq 0$. Then if $$\lim_{n \to \infty} \frac{a_n}{b_n} = L > 0,$$ the series $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty b_n$ converges.

Examples

1. Does the series $$\sum_{n=153}^\infty \frac{1}{n^2}$$ converge?

Solution:
Yes! This series is a tail of the convergent $p$-series $$\sum_{n=1}^\infty \frac{1}{n^2}$$ in this case, $p = 2$.


2. Show that $$\sum_{n=0}^\infty {n^5\over 5^n}$$ converges.

Solution:
First, we can easily prove that $2^n > n^5$ when $n$ is sufficient large (by Mathematical Induction). Suppose that $2^n > n^5$, then when $n > 1$, we have $$(n+1)^5=n^5+5n^4+10n^3+10n^2+5n+1 < n^5+5n^4+10n^3+10n^2+6n$$ $$< n^5+5n^4+10n^3+16n^2 < n^5+5n^4+26n^3 < n^5+31n^4$$ And thus $(n+1)^5 < 2n^5 < 2\cdot 2^n=2^{n+1}$ whenever $n > 31$.
Actually we can find the minimal $n$ that $2^n > n^5$, R code:

f1 = function(x) 2^x
f2 = function(x) x^5
for (i in 2:100){
  if (f1(i) > f2(i)){
    print(i)
    break
  }
}
## [1] 23

(Note that when $n=2, 3, \cdots$, $2^n < n^5$, thus we search from $n=2$.) Back to this problem, we have $${n^5 \over 5^n} < {2^n \over 5^n}=({2\over5})^n$$ That is, the tail series $$\sum_{n=23}^{\infty}({2\over5})^n$$ converges. By the comparison test, the smalled series $$\sum_{n=23}^{\infty}{n^5 \over 5^n}$$ also converges, so does the original series $$\sum_{n=0}^\infty {n^5\over 5^n}$$
3. Does the series $$\sum_{n=52}^\infty \frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2}$$ converges or diverge?

Solution:
By limit comparison test, set $$a_n=\frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2},\ b_n={1\over n}$$ We have $$\lim_{n\to\infty}{a_n\over b_n}=\lim_{n\to\infty}{\frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2}\over{1\over n}}=\lim_{n\to\infty}{n^5-3n^2+5n\over2n^5+5n^3-n^2}=\frac{1}{2} > 0$$ Thus, $\sum a_n$ and $\sum b_n$ share the same fate. But $b_n$ is harmonic series which diverges. Hence, $$\sum_{n=52}^\infty \frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2}$$ diverges.