Can i prove in this way?(convergence of infinite sum )

Question

The infinite sum written as $\sum_{k=1}^\infty a_k$ converges if and only if $|\sum_{k=n}^m a_k | < \varepsilon$ holds for all $m > n > N$. for all $\varepsilon$>0

The infinite sum converges: $|s_{n-1} - L| < \varepsilon$, for $n-1 > N$, where $s_n$ is $n$-th partial sum of complex numbers $(a_n)$, iff $(s_{n- 1})$ is Cauchy iff $|s_m - s_{n-1}| < \varepsilon$, for $m > n > n-1 > N$ equals to $$ \left| \sum_{k=n}^ma_k \right| < \varepsilon $$ hence proved the claim

Answer 1

What you've written doesn't appear to make sense. Here's how I would do it:

Let $(s_n)$ be the sequence of partial sums. Then $\sum a_n$ converges iff $(s_n)$ converges iff $(s_n)$ is Cauchy iff for all $\epsilon>0$ there's an $N$ such that whenever $n,m \geq N$ we have $\left|s_n-s_m \right|=\left|\sum_{k=m+1}^{n} a_k \right|<\epsilon$.

Answer 2

You've written "The infinite sum written as $\sum_{k=1}^\infty a_k$ converges if and only if $|\sum_{k=m}^n a_k|<\epsilon$ holds for all $m>n>N$. for all $\epsilon>0$."

This is understandable but it would be better to say "if and only if given $\epsilon>0$ we can find an $N$ such that $|\sum_{k=m}^n a_k|<\epsilon$ holds for all $m>n>N$."

Your proof is "$|s_{n−1}−L|<\epsilon$, for $n−1>N$, where $s_n$ is $n$-th partial sum of complex numbers $a_n$, iff $s_{n−1}$ is Cauchy iff $|s_m−s_{n−1}|<\epsilon$, for $m>n>n−1>N$, equals to $\left| \sum_{k=n}^ma_k \right| < \epsilon$."

This would be clearer if you said simply "$s_n$ tends to a limit, where $s_n$ is $n$-th partial sum of complex numbers $a_n$, iff $s_n$ is Cauchy iff given $\epsilon>0$ there is $N$ such that $|s_m-s_{n-1}|<\epsilon$ for $m>n>N$ which in turn is equivalent to $\left| \sum_{k=n}^ma_k \right| < \epsilon$ for $m>n>N$."

There is nothing in what you wrote which is wrong mathematically, but it is not expressed as well as it might be. In particular,

(a) in English facts are not called equal to other facts but are instead called equivalent to those facts, and

(b) it's more legible to say expressly things like "given $\epsilon>0$ there exists some $N$" which satisfies a condition. If you introduce $\epsilon$ and $N$ without explanation your statement will usually be intelligible (or guessable), but why take the risk that it won't be? Plus it's good to make things easier on your readers.

Answer 3

To understand what is going on it is helpful to approach such a problem in two stages: (1) understanding, and (2) formalisation and epsilon-N paraphrase.

At stage (1) the point is that for infinite $H$ the remainder term $r_H=\sum_{k=H+1}^\infty a_k$ is infinitesimal.

Once you have an intuitive grasp of the concept, go on to the formalisation stage in terms of the following long-winded paraphrase, whose effect is to replace infinite and infinitesimal by very large and very small respectively, while paying attention to the quantifier order (which can be tricky) namely:

For every $\epsilon>0$ there exists an $N$ such that if $n,m>N$ then the truncated partial sum $\sum_{k=n}^{k=m}a_k$ will be smaller than $\epsilon$ in absolute value.