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Suppose that the series $\sum a_n$ converges. Prove that $\lim\limits_{n\to\infty} a_n = 0$.

Not sure how to do this my attempt:

WTS:$\forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|a_n - 0| < \epsilon$

Let $\epsilon > 0$ be arbitrary

Choose N such that $|a_n - 0| < \epsilon$

Suppose $n > N$, then $|a_n - 0| < \epsilon$

I am not sure.

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  • $\begingroup$ I changed it, thank you. $\endgroup$ – user349557 Apr 19 '17 at 5:57
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    $\begingroup$ The sequence of partial sums $S_n$ converges, hence the sequence $\{S_n\}$ is Cauchy. This implies for any $\epsilon > 0$ and sufficiently large $n$, we have $|S_{n+1} - S_n| < \epsilon$ Since $S_{n+1} - S_n = a_{n+1}$, the result follows $\endgroup$ – MathematicsStudent1122 Apr 19 '17 at 6:02
  • $\begingroup$ This is an excellent solution, however, from seeing the OPs work; I wonder if he knows how to show a convergent sequence is Cauchy..... $\endgroup$ – ADA Apr 19 '17 at 6:09
  • $\begingroup$ > I had to google what cauchy was. $\endgroup$ – user349557 Apr 19 '17 at 6:27
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Let $A_n := \displaystyle \sum_{k=1}^n a_k$. Note that $a_n = A_n - A_{n-1}$.

Then, $\displaystyle \sum_{k=1}^\infty a_k := \lim_{n\to\infty} A_n = L$.

Now, since the sum converges:

$$\forall \varepsilon_1 > 0: \exists N_1 \in \Bbb N: \forall n > N_1:|A_n-L| < \varepsilon_1$$

Now, we need to prove that $a_n$ goes to zero:

$$\forall \varepsilon_2 > 0: \exists N_2 \in \Bbb N: \forall n > N_2:|a_n| < \varepsilon_2$$

To prove that, let $\varepsilon_1 = \dfrac12\varepsilon_2$. Now, we have $N_1$ from the assumption. Let $N_2 = N_1+1$. For any $n > N_2$:

$$\begin{array}{rcll} |A_{n+1} - L| &<& \dfrac12 \varepsilon_2 & \text{assumption} \\ |A_n - L| &<& \dfrac12 \varepsilon_2 & \text{assumption} \\ |A_{n+1} - A_n| &<& |A_{n+1} - L| + |A_n - L| & \text{triangle inequality} \\ &<& \dfrac12 \varepsilon_2 + \dfrac12 \varepsilon_2 \\ &=& \varepsilon_2 \\ |a_n| &<& \varepsilon_2 & \text{conclusion} \end{array}$$

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Suppose that $\sum\limits_{n=1}^{\infty}{a_n}=M$ then we have that $M=\lim_{n \to \infty}({a_n} + \sum\limits_{k=1}^{n-1}{a_k})=\lim_{n\to\infty}{a_n}+\lim_{n\to\infty}{\sum\limits_{k=1}^{n-1}{a_n}}=\lim_{n\to\infty}a_n+M$. Hence, we have that $a_n \to 0$.

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  • $\begingroup$ good answer.a new proof! $\endgroup$ – vidyarthi Apr 19 '17 at 6:08
  • $\begingroup$ Glad you like it! $\endgroup$ – ADA Apr 19 '17 at 19:13
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Let $$L=\sum^{\infty}_{n=1}a_{n}=\lim _{n\to \infty}\sum^{n}_{k=1}a_{k}=L$$ Then, $A_{n}$ is from Cauchy, that is: $$\forall \epsilon >0, \exists n_{0}\in \mathbb{N}/ n,p\geq n_{0}\rightarrow |A_{p}-A_{n}|<\epsilon$$ In particular, for $$p=n+1\rightarrow |A_{n+1}-A_{n}|<\epsilon \rightarrow \left | \sum^{n+1}_{k=1}a_{k}-\sum^{n}_{k=1}a_{k}\right|<\epsilon \rightarrow |a_{n+1}|<\epsilon\rightarrow$$ Taking $$n_{1}=n_{0}+1\rightarrow \forall \epsilon >0, \exists n_{0}\in \mathbb{N}/ n\geq n_{1}\rightarrow |a_{n}-0|<\epsilon \rightarrow \lim _{n\to \infty}a_{n}=0$$

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