# 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}