Sol Verif: Prove that if $\lim_{n \rightarrow \infty }\sum u_n=l\Rightarrow \lim_{n \rightarrow \infty} u_n=0$ via Cauchy

Question

Question:
Prove that if $\lim_{n \rightarrow \infty }\sum u_n=l\Rightarrow \lim_{n \rightarrow \infty} u_n=0$ by using Cauchy definition of convergence.

My answer:
1- First of all let's write: $S_n=\sum_{k=1}^{n} u_k$. As it is given by assumption we have that $\lim_{n \rightarrow \infty }S_n=l$ so $S_n$ converges and is a Cauchy sequence.

2- Now let's choose one $\epsilon_0>0$ and WLOG we choose $p>q$ and so $S_p-S_q=\sum_{k=q+1}^{p} u_k$.
By definition of a Cauchy sequence it exists at least one $N_{\epsilon_0} \in \mathbb{N} $ s.t. $\forall p,q>N_{\epsilon_0} \Rightarrow |S_p-S_q|<\epsilon_0$. Now if it is true $\forall p,q>N_{\epsilon_0}$ it is in particular true for the couple $\left (p;q \right )$ choosed as follow $\left ( q, p=q+1 \right )$ and for the couple $\left ( q+1; p=q+2 \right )$...etc... In such cases we get $|S_p-S_q|=|u_p|<\epsilon_0$. In other words for the given $\epsilon_0>0 $, $\exists N_{\epsilon_0} \in \mathbb{N}$ s.t. we have an infinity element of the form $u_{q+k}$ in the interval $\left ( -\epsilon_0, \epsilon_0 \right )$ with $q>N_{\epsilon_0}, k \in \mathbb{N}$.

3- The same reasoning is valid $\forall \epsilon>0$ (trivial).
Hence it means that $\forall \epsilon>0$ we will always be able to find a $ N_{\epsilon} \in \mathbb{N}$ such that there will be an infinite amount element of the form $u_{q+k}, k \in \mathbb{N}$ in the interval $\left ( -\epsilon, \epsilon \right )$ at the condition that $q> N_{\epsilon}$. And this is exactly the definition of $\lim_{n \rightarrow \infty} u_n=0$.

Q.E.D.

Summarize, my prove is base on the fact that because $S_n$ is a Cauchy sequence, by choosing wisely $p,q>N$ of the definition we can see that the Cauchy definition defines too a sequence $u_n$ that converges to zero. I hope it is clear enough.

Is it correct?

Answer 1

In order to prevent the following confusion undersigned by @Andrew that my previous prove: " suggests that you do not really understand limits."[...] " that there is a subsequence which converges, not the entire sequence.".

I think I can be more precise and re-write my proof.

1- First of all let's write: $S_n=\sum_{k=1}^{n} u_k$. As it is given by assumption we have that $\lim_{n \rightarrow \infty }S_n=l$ so $S_n$ converges and is a Cauchy sequence.

2- Now let's choose one $\epsilon_0>0$ and WLOG we choose $p>q$ and so $S_p-S_q=\sum_{k=q+1}^{p} u_k$.
By definition of a Cauchy sequence it exists at least one $N_{\epsilon_0} \in \mathbb{N} $ s.t. $\forall p,q>N_{\epsilon_0} \Rightarrow |S_p-S_q|<\epsilon_0$. Now if it is true $\forall p,q>N_{\epsilon_0}$ it is in particular true for all the couples $\left (p;q \right )$ choosed as follow $\left ( q, p=q+1 \right )$ and for the couple $\left ( q+1; p=q+2 \right )$...etc... In such cases we get $|S_p-S_q|=|u_p|<\epsilon_0$. In other words for the given $\epsilon_0>0 $, $\exists q_{\epsilon_0} > N_{\epsilon_0} \in \mathbb{N}$ s.t. $\forall k \in \mathbb{N} \Rightarrow u_{q+k} \in \left( -\epsilon_0, \epsilon_0 \right )$.
In words it means that a direct consequence of the Cauchy definition is that we can wisely choose couples $\left( p;q \right )$ s.t. we will get that all the elements of the form/sequence $u_{q+k}$ will be in the interval $\left( -\epsilon_0, \epsilon_0 \right )$

3- The same reasoning is valid $\forall \epsilon>0$ (trivial). So we get exactly the definition of $\lim_{n \rightarrow \infty} u_n=0$.

Q.E.D.

Is it better? Correct?

Answer 2

It seems that my first answer was correct.

Btw here an other and more simple proof that does not use Cauchy (in the case someone will need it in the futur).

By assumption we have $\lim_{n \to \infty } S_n=l$, hence all the sub sequences of $S_n$ converges to $l$.
So in particular for the sub sequence of the form: $S_{n-1}$, $\lim_{n \to \infty } S_{n-1}=l$.
Now we note that we have $S_n-S_{n-1} = u_n$ and by limit arithmetic we get:$ S_n-S_{n-1} = u_n \underset{n \to \infty }{\rightarrow}l-l=0 $.

Q.E.D.