# Why does the tail $a_N+a_{N+1}+a_{N+2}\ldots$ of convergent series $\sum a_n$ tend to $0$ as $N\to\infty$?

Let $a(n)>0$ for all $n \in \mathbb{N}$ be such that $\sum a(n)$ converges. Define $r(n):=\sum\limits_{k=n+1}^\infty a(k)$. The claim is $$\lim_{n \to \infty} r(n) = 0,$$ but I cannot see how to prove it. Please help.

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I have changed some things, are they still correct as per your question? – Alex J Best Sep 7 '12 at 21:27
It is perfect , but I cannot find any change – Ester Sep 7 '12 at 21:29
He set the math in $\LaTeX$ and wanted to be sure he got it right. If you want to do your own, you could see meta.math.stackexchange.com/questions/1773/… for some hints. – Ross Millikan Sep 7 '12 at 21:48
What about if the series is divergent ? – marios Apr 22 at 23:57

If the series $a(n)$ converges to $A$, that means that the sums $\sum_{i=0}^n a(i)$ get very close to $A$ as $n$ gets large. You have $r(n)=\sum_{j=n+1}^\infty a(j)=A-\sum_{i=0}^n a(i)$ and the right side goes to $0$ as $n$ gets large.

To make it more formal, you can use an $\epsilon N$ argument: Given $\epsilon \gt 0$ we need to find $N$ such that there is an $N$ such that $r(n)=\sum_{j=N+1}^\infty a(j) \lt \epsilon$. We are told that there is an $N$ such that $A-\sum_{i=0}^N a(i)\lt \epsilon$ and the same $N$ works.

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Thanks to all of you for your kind help. – Ester Sep 7 '12 at 21:44

We know that the series $$\sum_{k=1}^\infty a(k)$$ converges, meaning by definition that its sequence of partial sums $$s(n):=\sum_{k=1}^na(k)$$ converges. Take $\varepsilon>0$, so by convergence of its sequence of partial sums, we have for some $N$ and all $n\geq N$ that $$\left|s(n)-\sum_{k=1}^\infty a(k)\right|<\varepsilon.$$ But $$s(n)-\sum_{k=1}^\infty a(k)=-r(n),$$ so $-r(n)\to 0$, and so $r(n)\to 0$.

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So $\lim r(n) = 0$ for a positive sequence $r(n)$ means that for every positive $\epsilon$ we can find an $N$ where $m > N$ implies $r(m) < \epsilon$.

On the other hand, the (strictly increasing) series $a(n)$ converging to, say, $\alpha$ means that for every positive $\delta$, we can find a $M$ where $p > M$ implies that $\alpha - \delta < \sum_{i=1}^p a(i) < \alpha$.

Now rewrite $\alpha$ in terms of the series $a(n)$,

$\sum_{i=1}^\infty a(i) -\delta < \sum_{i=1}^m a(i) < \sum_{i=1}^\infty a(i)$

Subtract off the partial sequence and add $\delta$,

$\sum_{i=m+1}^\infty a(i) < \delta$

Since $\delta$ was arbitrary, and the left hand side positive, we know the limit is zero.

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Let $s(n) = \sum _{k = 0}^n a(k)$. Let $s = \sum _{k = 0}^\infty a(k)$.

Then $\sum _{k = n+1}^\infty a(k) = s - s(n)$.

Hence $\lim_{n\rightarrow\infty} \sum _{k = n+1}^\infty a(k) = \lim_{n\rightarrow\infty} s - s(n) = 0$

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