Multiplying a divergent series by a constant doesn't affect the series divergence?

Question

Is this true? I'm unsure as of right now if this is the case, and I can't find any reference to this on Stack exchange after about 10 minutes of searching; if this has been asked before I really do apologise!

Answer 1

Let $\sum a_n$ be a non convergent series. By definition, this means that the sequence $s_n := \sum_{j=1}^n a_j$ of partial sums does not converge.

If $c\neq 0$, then the partial sums $(\sigma_n)$ of the series $\sum (c a_n)$ are given by $$ \sigma_n = \sum_{j=1}^n (c a_j) = c \sum_{j=1}^n a_j = c\, s_n. $$ Since $(s_n)$ does not converge and $c\neq 0$, also $(\sigma_n)$ does not converge.

Answer 2

$\{a_n\}$ converging to $L$ means for any $\epsilon > 0$ there is an $N$ so that $n > N$ implies $|a_n - L| < \epsilon$.

$\{a_n\}$ diverges means, depending upon your definition either, $\{a_n\}$ does not converge or for any $M$ there is an $N$ so that $n > N$ implies $|a_n| > M$.

If $k \ne 0$ and if $\{a_n\}\rightarrow L$ then for any $\epsilon > 0$ there is and $\epsilon/|k| > 0$ so there is an $N$ so that $n > N$ implies $|a_n -L| < \epsilon/|k|$ so $|k*a_n - k*L| = |k|*|a_n - L| < \epsilon$. And as $k \ne 0 \iff \frac 1k \ne 0$. $\{a_n\} \rightarrow L \iff \{ka_n\} \rightarrow kL$.

If $k \ne 0$ and if $\{a_n\}\rightarrow \infty$ then for any $M$ there is an $M/|k|$ and there is an $N$ so that $n > N$ implies $|a_n| > M/|k|$ and $|ka_n| > M$ so $\{a_n\} \rightarrow \infty \iff \{ka_n\} \rightarrow \infty$

A series $\sum^{\infty} b_n = \lim a_n$ where $a_n = \sum^n b_n$.