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Suppose that $\sum_{k=1}^{\infty}a_k$ is a convergent series.

a) Prove that the series $\sum_{k=n}^{\infty}a_k$ converges for each positive integer n.

b)Let {$t_n$} = $\sum_{k=n}^{\infty}a_k$. Prove that the sequence {$t_n$} converges to 0.

So I am done with part a and understand it. But I am having a hard time conceptualizing b. Would the terms of that series be decreasing? Any help would be appreciated. Having a hard time figuring it out.

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Since $\sum a_k$ is convergent the sequence $s_n=\sum_{k=1}^na_k$ of partial sums is convergent and hence Cauchy. This means that:

$$\forall \epsilon>0,\,\exists n_0 \in \mathbb{N},\,\forall m,n\geq n_0,\,|s_m-s_n|\leq \epsilon$$

Notice that if $m>n$, then $s_m-s_n =\sum_{k=n+1}^ma_n$. Letting $m\to\infty$, we concude that

$$\forall \epsilon>0,\,\exists n_0 \in \mathbb{N},\,\forall n\geq n_0,\,\left|\sum_{k=n+1}^\infty a_n\right|\leq \epsilon.$$

Notice that this step is allowed because of part (a).
It suffices to note that the statement above means precisely that $t_n\to0$.

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  • $\begingroup$ How does it mean that $t_n$ goes to 0? Sorry could you please elaborate on that I am having a hard time piecing it together $\endgroup$ – JxxYsde3 Feb 15 '17 at 4:32
  • $\begingroup$ Technically, it means $t_{n+1}=\sum_{k=n+1}^\infty a_n$ goes to $0$ (as $n\to\infty$). This is by the definition of limits. $\endgroup$ – Fimpellizieri Feb 15 '17 at 4:36
  • $\begingroup$ I'm sorry, I'm just having a difficult time understanding this --- how does this prove that it converges to 0? $\endgroup$ – JxxYsde3 Feb 15 '17 at 4:40
  • $\begingroup$ By definition. You should lookup the definition of what it means for some sequence $x_n$ to converge to some limit $L$. $\endgroup$ – Fimpellizieri Feb 15 '17 at 5:17

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