Answer
$\Sigma^{\infty}_{n=1}Ca_{n}=C\Sigma^{\infty}_{n=1}a_{n}$
Work Step by Step
Prove of Theorem 8 (i): $\Sigma^{\infty}_{n=1}Ca_{n}=C\Sigma^{\infty}_{n=1}a_{n}$
If $\Sigma a_{n}$ is convergent, then we have to prove that $\Sigma^{\infty}_{n=1}Ca_{n} = \Sigma^{\infty}_{n=1}a_{n}$, where $C$ is constant.
Since the series is convergent, $\Sigma^{\infty}_{n=1}a_{n}$ exists and is finite.
$\Sigma^{\infty}_{n=1}Ca_{n} = \lim\limits_{n \to \infty}\Sigma^{n}_{i=1}Ca_{i}$
$=\lim\limits_{n \to \infty}C\Sigma^{n}_{i=1}a_{i}$
$=C\lim\limits_{n \to \infty}\Sigma^{n}_{i=1}a_{i}$
$=C\Sigma^{\infty}_{n=1}a_{n}$
Since $\Sigma^{\infty}_{n=1}a_{n}$ exists and is finite, then $C\Sigma^{\infty}_{n=1}a_{n}$ will also exist and is finite.
Hence, the theorem is proved.
$\Sigma^{\infty}_{n=1}Ca_{n}=C\Sigma^{\infty}_{n=1}a_{n}$
