9.3 | Convergence of Series


Convergence Properties of Series:

  1. If \(\displaystyle \sum_{n=1}^\infty a_n \) and \(\displaystyle \sum_{n=1}^\infty b_n \) converge and if \(k\) is a constant, then
    • \(\displaystyle \sum_{n=1}^\infty (a_n+b_n) \) converges and is equal to \(\displaystyle \sum_{n=1}^\infty a_n + \displaystyle \sum_{n=1}^\infty b_n \).
    • \(\displaystyle \sum_{n=1}^\infty k a_n \) converges to \(k \displaystyle \sum_{n=1}^\infty a_n \).
  2. Changing a finite number of terms in a series does not change whether or not it converges, although it may change the value of its sum if it does converge.
  3. If \(\displaystyle \lim_{n \rightarrow \infty} a_n \not = 0\) or \(\displaystyle \lim_{n \rightarrow \infty} a_n\) does not exist, then \(\displaystyle \sum_{n = 1}^\infty a_n\) diverges.
  4. If \(\displaystyle \sum_{n = 1}^\infty a_n\) diverges, then \(\displaystyle \sum_{n = 1}^\infty k a_n\) diverges if \(k \not = 0\).
The Integral Test: Let \(f(x)\) be a positive, continuous, decreasing function on \([1, \infty) \). Then
$$
\sum_{n = 1}^{\infty} a_n \text{ and } \int_{1}^{\infty} f(x) \; dx
$$
either both converge or both diverge.
A series which can be written in the form
$$
\sum_{n=1}^{\infty} \frac{1}{n^p}
$$
is called a \(p\)-series and
$$
\sum_{n=1}^{\infty} \frac{1}{n^p} =
\begin{cases}
\text{convergent}, & \; \text{ if } p > 1 \\
\text{divergent}, & \; \text{ if } p \leq 1
\end{cases}
$$
A series which can be written in the form
$$
\sum_{n = 1}^{\infty} \frac{1}{n}
$$
is called a harmonic series and diverges.
Determine whether the series is convergent or divergent.
$$
\sum_{n=1}^{\infty} \left(\frac{-6}{\sqrt{n^3}} + \frac{-5}{n^5}\right)
$$
If the series is convergent, find the 4th partial sum.


$$
\sum_{n=1}^{\infty} \left(\frac{-6}{\sqrt{n^3}} + \frac{-5}{n^5}\right)
= -6\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}}_{\text{convergent p-series}} -5\underbrace{\sum_{n=1}^{\infty} \frac{1}{n^5}}_{\text{convergent p-series}}
$$
$$
\begin{aligned}
\sum_{n=1}^{\infty} \left(\frac{-6}{\sqrt{n^3}} + \frac{-5}{n^5}\right)
& = -6\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}}_{\text{convergent p-series}} \\
& \; \; -5\underbrace{\sum_{n=1}^{\infty} \frac{1}{n^5}}_{\text{convergent p-series}}
\end{aligned}
$$
$$
\begin{aligned}
\sum_{n=1}^{\infty} \left(\frac{-6}{\sqrt{n^3}} + \frac{-5}{n^5}\right)
& = -6\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}}_{\text{convergent p-series}} \\
& \; \; -5\underbrace{\sum_{n=1}^{\infty} \frac{1}{n^5}}_{\text{convergent p-series}}
\end{aligned}
$$

A constant multiple of a convergent series is convergent and the sum or difference of two convergent series is convergent. So, the given series is convergent.

The 4th partial sum is given by:
$$
\begin{aligned}
s_4 & = \sum_{n=1}^4 \left(\frac{-6}{\sqrt{n^3}} + \frac{-5}{n^5}\right) \\
& = \overbrace{-6 -5}^{n=1} + \overbrace{\frac{-6}{\sqrt{2^3}} + \frac{-5}{2^5}}^{n=2} \\
& \; \; + \overbrace{\frac{-6}{\sqrt{3^3}} + \frac{-5}{3^5}}^{n=3} + \overbrace{\frac{-6}{\sqrt{4^3}} + \frac{-5}{4^5}}^{n=4} \\
& = -\frac{2968991}{248832} – \frac{3}{\sqrt{2}} – \frac{2}{\sqrt{3}}
\end{aligned}
$$

E 9.3 Exercises