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Given a convergent series

$$\sum\limits_{k = 1}^\infty a_k < \infty$$

Then claim:

$\exists N$, such that $\exists \epsilon > 0$, $\sum\limits_{k = > N+1}^\infty a_k < \epsilon$

I tried to go by definition

$$\sum\limits_{k = 1}^\infty a_k = L < \infty$$ is convergent if $\forall \epsilon > 0, \forall k \geq N, |L - \sum\limits_{k = 1}^\infty a_k | < \epsilon$

But the claim eliminates the $L$...not sure what to do next

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    $\begingroup$ Let $N=0$ and $\varepsilon=\left|\sum \limits_{k=1}^\infty\left(a_k\right)\right|+1$. $\endgroup$ – Git Gud Mar 26 '16 at 22:19
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    $\begingroup$ You probably want to show that for all $\epsilon \gt 0$ there is an $N$ such that $\dots$. $\endgroup$ – André Nicolas Mar 26 '16 at 22:25
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You have a little mistake in your argument. What $\displaystyle{\sum_{k=1}^\infty a_k=L<\infty}$ means is that the sequence of partial sums $\{\displaystyle{S_n=\sum_{k=1}^n a_k}:n\in\mathbb{N}\}$ converges to $L$, that is, for all $\epsilon>0$ there exists $N\in \mathbb{N}$ such that if $n\geq N$ we have $|S_n-L|<\epsilon$, that is:

$$\forall \epsilon>0 \left(\exists N\in\mathbb{N}\left(n\geq N \rightarrow |L-S_n|=\left|\sum_{k=1}^\infty a_k - \sum_{k=1}^n a_k\right|<\epsilon\right)\right)$$

In particular, for $n=N$ we have:

$$\sum_{k=1}^\infty a_k-\sum_{k=1}^Na_k=\sum_{k=N+1}^\infty a_k<\epsilon ,$$as desired.

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  • $\begingroup$ Good job although crazy amount of brackets for your definition where you learn that $\endgroup$ – Shamisen Expert Mar 27 '16 at 0:02
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    $\begingroup$ Indeed, much too many brackets, especially in view of the missing "$\forall n\in\mathbb N$". $\endgroup$ – Did Mar 27 '16 at 7:36
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Your statement is wrong.

$\sum_{k=1}^\infty a_k=L$ means that the sequence of partial sums $s_n=\sum_1^n a_k$ converges to $L$.

That is, for any $\varepsilon>0$, there exists $N$ such that $|L-s_n|<\varepsilon\forall n>N$

However, $L-s_n=\sum_{1}^\infty a_k-\sum_1^n a_k=\sum_{N+1}^\infty a_k$.

Thus, you reach the needed conclusion.

However, if one needs strict rigour regarding the sum from $N+1$ to $\infty$, then one might note:

$s_n-s_m=\sum_{n+1}^m a_k$ for all $m>n$.

and then take the limit of $m$ as it goes to $\infty$. to see that $L-s_n=\sum_{N+1}^\infty a_k$

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  • $\begingroup$ Actually the statement in the question is not wrong, it always hold. $\endgroup$ – Did Mar 26 '16 at 22:30
  • $\begingroup$ Not when he says that he's going by the definition ;) (I meant that his statement that this is the definition is wrong) $\endgroup$ – Hasan Saad Mar 26 '16 at 22:31
  • $\begingroup$ Oh, you mean "I tried to go by definition $\sum\limits_{k = 1}^\infty a_k = L < \infty$ is convergent if $\forall \epsilon > 0$, $\forall k \geq N$, $\left|L - \sum\limits_{k = 1}^\infty a_k \right| < \epsilon$"? Right, this part escaped me. (Dunno whether it is more wrong than absurd or the opposite, though...) $\endgroup$ – Did Mar 26 '16 at 22:34

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