这是一份nottingham诺丁汉大学MATH1006作业代写的成功案例
Given a series $\Sigma_{m=1}^{m} a_{n}=a_{1}+a_{2}+a_{s}+\cdots$, let $s_{a}$ denote its rth partial sum:
$$
s_{n}=\sum_{i=1}^{n} a_{i}=a_{1}+a_{2}+\cdots+a_{n}
$$
If the sequence $\left{s_{\mathrm{n}}\right}$ is convergent and $\lim {\mathrm{a} \rightarrow \mathrm{m}} s{\mathrm{a}}=s$ exists as a real number, then the series $\Sigma a_{n}$ is called convergent and we write
$$
a_{1}+a_{2}+\cdots+a_{n}+\cdots=s \quad \text { or } \quad \sum_{n=1}^{\infty} a_{n}=s
$$
The number $s$ is called the sum of the series. Otherwise, the series is called divergent.
MATH1006 COURSE NOTES ：
If $r=1$, then $s_{n}=a+a+\cdots+a=n a \rightarrow \pm \infty$. Since $\lim {n \rightarrow-} s{n}$ doesn’t exist, the geometric series diverges in this case.
If $r \neq 1$, we have
$$
\begin{aligned}
&s_{\mathrm{a}}=a+a r+a r^{2}+\cdots+a r^{\mathrm{n}-1} \
&r s_{\mathrm{a}}=\quad a r+a r^{2}+\cdots+a r^{\mathrm{n}-1}+a r^{\mathrm{n}}
\end{aligned}
$$
Subtracting these equations, we get
$$
\begin{array}{r}
s_{\mathrm{a}}-r s_{\mathrm{a}}=a-a r^{n} \
s_{\mathrm{n}}=\frac{a\left(1-r^{\mathrm{n}}\right)}{1-r}
\end{array}
$$
If $-1<r<1$, we know from $(11.1 .9)$ that $r^{n} \rightarrow 0$ as $n \rightarrow \infty$, so
$$
\lim {n \rightarrow \infty} s{n}=\lim {n \rightarrow \infty} \frac{a\left(1-r^{n}\right)}{1-r}=\frac{a}{1-r}-\frac{a}{1-r} \lim {n \rightarrow \infty} r^{n}=\frac{a}{1-r}
$$