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I just had this problem on a final and I was confused about proving the reverse direction. For example, what if $a_n = \frac{1}{(n-1)^2}$?

Then $\sum_{n=100}^{\infty} a_n$ would converge, but the first term in $\sum_{n=1}^{\infty} a_n$ would be undefined.

I think I'm missing something. Any help would be appreciated.

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  • $\begingroup$ Well, $a_n$ is a real sequence, so it is a map from the natural numbers to the real numbers. But then, $a_1$ does not map to the real numbers at all, so $a_n$ isn't even a real sequence, simply because $a_1$ is flouting the norms. $\endgroup$ – астон вілла олоф мэллбэрг Dec 13 '17 at 5:12
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    $\begingroup$ You are probably expected to assume that $a_n$ is defined for all $n\ge1$. Was there maybe an introduction to the question saying something like "for all $n\ge1$, let $a_n$ be a real number"? $\endgroup$ – David Dec 13 '17 at 5:13
  • $\begingroup$ It seems in this theorem one must assume $a_n$ is defined for all $n \ge 1,$ otherwise the left sum is not defined. $\endgroup$ – coffeemath Dec 13 '17 at 5:14
  • $\begingroup$ I suppose $a_n$ is defined for all natural number $n$. $\endgroup$ – Idonknow Dec 13 '17 at 5:14
  • $\begingroup$ @David It was not specified that $a_n$ must be defined for all $n \geq 1$ or anything along those lines... I just chose to pretend that was given when doing the proof and hoped for the best with my test score. $\endgroup$ – cedge113 Dec 13 '17 at 5:18
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For $n \in \mathbb N$ let $S_n=\sum_{k=1}^na_k$ and for $n \ge 100$ let $s_n=\sum_{k=100}^na_k$. Furthermore let$a:=a_1+...+a_{99}$.

If $n \ge 100$ we then have

$S_n=a+s_n$.

We get:

$\sum_{n=1}^{\infty} a_n$ is convergent $ \iff (S_n)$ is convergent $ \iff (s_n)$ is convergent $ \iff \sum_{n=100}^{\infty} a_n$ is convergent.

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Cauchy criterion:

$S_m = \sum_{i=100}^{m} a_k$ convergent .

Let $\epsilon \gt 0$ be given.

There is a $m \ge n \ge n_0 \gt 100$ such that

$|S_m-S_n| \lt \epsilon.$

Use this $n_0$ to establish that

$R_m= \sum_{i=1}^{m} a_i$ is Cauchy,

hence convergent.

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