How can we formally state a series converges?

Question

If the series $\sum_{n=0}^{\infty}na_{n}$ converges absolutely, how can we write this using the definition of convergence?

The definition of convergence states this:

"The series $\sum_{n=1}^{\infty}a_{n}$ converges if and only if, given $\epsilon \gt 0$, there exists an $N \in \mathbf N$ such that whenever $n \gt m \ge N $ it follows that: |$ a_{m+1} + a_{m + 2} + ... + a_{n}| \lt \epsilon$ "

How do I extend this to my current series above? What is tricking me and I'm not too sure if what I'm doing is correct is the $n$ infront of the $a_{n}$. If that series converges absolutely, does it imply this:

|$ na_{m+1} + na_{m + 2} + ... + na_{n}| \lt \epsilon$ ?

Any help would be appreciated.

Answer 1

The correct inequality is $(m+1)|a_{m+1}|+(m+2)|a_{m+2}|+\cdots+n|a_n|<\epsilon$.

Answer 2

The series $\sum_{k=0}^\infty b_k$ converges absolutely $\iff$ The series $\sum_{k=0}^\infty |b_k|$ converges.

In your case, take $b_k = ka_k$ and plug it into whatever your definition is for (series) convergence.

Usually, the definition of (series) convergence is the following: The series $\sum_{k=0}^\infty b_k$ converges to a number $L$ $\iff$ "For all $\epsilon > 0$, there exists an $N > 0$ such that $n \geq N$ implies $\left| \sum_{k=1}^n b_k - L \right| < \epsilon$."

As Henning Makholm correctly said in the comments, the definition you gave is for a series being Cauchy. In the case where the terms of your series are all real numbers, a series being Cauchy is equivalent to a series being convergent.