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So let $\lim_{n\rightarrow\infty}a_n = a$ and for simplicity let $\{a_n\}$ be increasing and $a>0$. Now consider $\sum_n (a- a_n)$. I wrote a proof that this doesn't converge, and I'm wondering where I went wrong, or if it just works because we have an increasing sequence (part of the proof relies on it being monotone).

Proof: For any $\epsilon > 0$ we have that there exists an $N$ so that the terms of the series $a-a_n = b_n <\epsilon$ for all $n>N$. In particular consider $\epsilon$ of the form $\frac{1}{k}$ for some integer $k$. We have that for every $k\geq 1$ that there exists an $N$ so that $b_n < \frac{1}{k}$ for all $n>N$.

Now since $\{a_n\}$ is increasing, we know that $b_n$ is bounded above by $a- a_0$. Let $k_0$ be the floor of $\frac{1}{a-a_0}$. Then for every integer $k>k_0$, there exists an integer $M$ so that for all $n\leq M$ we have that $b_n \geq \frac{1}{k}$.

Thus we have $\sum b_n \geq \sum_{k=k_0+1}^\infty \alpha_k\frac{1}{k}$ where the $\alpha_k$ are integer weights. They represent the distance between the $M$ we obtain for each consecutive $k$. In particular, $\alpha_k \geq 1$ for all $k$. Thus we have $\sum b_n \geq \sum_{k=k_0+1}^\infty \frac{1}{k}$ which proves divergence by the comparison test.

This completes the proof. Hopefully this is easy to read!

edit: seems that the critical problem is where I assumed that $\alpha_k \geq 1$ for all $k$. Since $\alpha_k$ = $M_k - M_{k-1}$ (where these $M$ come from paragraph 3) it could easily be the case that $\alpha_k = 0$, if the sequence approaches its limit fast enough.

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    $\begingroup$ Well, this statement is untrue when $a_n=1-\frac{1}{2^n}$, so try to figure out why this fails there. $\endgroup$ – Thomas Andrews Feb 7 '18 at 20:09
  • $\begingroup$ You want $a_n=1-\frac{1}{n}$ as your second example, to make it increasing. @lulu $\endgroup$ – Thomas Andrews Feb 7 '18 at 20:11
  • $\begingroup$ Of course, there is no reason to assume the $k$ is not the same for multiple $n.$ For your proof to work, you need $b_k\geq \frac{1}{k+k_0}.$ $\endgroup$ – Thomas Andrews Feb 7 '18 at 20:13
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    $\begingroup$ Basically, a lot of your $\alpha_n$ are zero. $\endgroup$ – Thomas Andrews Feb 7 '18 at 20:15
  • $\begingroup$ Yeah Thomas I think that's exactly the problem - $b_n$ could decrease so fast that many of the alphas are zero. $\endgroup$ – Ten Times Feb 7 '18 at 20:17
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The result itself is definitely not true: it can happen that $a_k$ converges rapidly enough that its deviations from its eventual limit are summable. For example this happens if $a_k=\sum_{n=1}^k 2^{-n}$.

The real error in your proof is that actually not all the $\alpha_k$ are $\geq 1$. For instance with this example $a_k$, the only $\alpha_k$ that are positive are those with $k$ given by a power of $2$.

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  • $\begingroup$ @DanielSchepler Oh, yes, we're comparing to the limit not just neighboring terms. Let me fix it... $\endgroup$ – Ian Feb 7 '18 at 20:17
  • $\begingroup$ I didn't mean to give a lower bound on $b_n$, but to say that $b_n$ is decreasing and that $a-a_0$ is its maximum, so if I choose a number smaller than $a-a_0$ and greater than $0$ then there will be a finite number of $b_n$ that are greater than it. $\endgroup$ – Ten Times Feb 7 '18 at 20:18
  • $\begingroup$ @TenTimes Your emphasis in your proof was backwards then: you wanted to say that $\{ n: b_n \geq 1/k \}=\{ 1,\dots,M(k) \}$. And that is true, but then it will happen that many of your $\alpha_k$ are in fact zero. For instance in my example, $\alpha_3=\alpha_5=\alpha_6=\alpha_7=0$. $\endgroup$ – Ian Feb 7 '18 at 20:20
  • $\begingroup$ Yeah I'm sorry; I need to get better at writing clearer; especially when I'm making up notation on the fly. Thanks for pointing this out. $\endgroup$ – Ten Times Feb 7 '18 at 20:23
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Let $$\beta_k=\left|\left\{n\mid b_n\geq \frac{1}{k}\right\}\right|$$

Then you are assuming that $\beta_k\geq k-k_0.$ Your assumption that $\alpha_k>0$ for each $k>k_0$ means that you assume that $\beta_k\geq k-k_0$ for each $k>k_0$. That's not going to happen when $b_n$ converges to $0$ very quickly.

For example, with $b_n=\frac{1}{2^n},$ you have that $\beta_k=\left\lfloor \log_2 k\right\rfloor$.

Now, $\beta_k$ is an increasing sequence of integers that converges to $+\infty$. If $k=O(\beta_k)$, then you get "enough" to ensure that the series $\sum b_n$ is bounded below by some constant times the harmonic series.

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On the contrary, every convergent series $\sum a_n $ enjoys the condition that $lim_{n\to \infty } a_n =0.$

Thus $\sum (a_n-a) = \sum (a_n-0) =\sum a_n $ has to converge.

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  • $\begingroup$ Those $a_n$ are not increasing and converging to a positive value. (You are correct that OP is wrong, but this answer is not helpful.) $\endgroup$ – Thomas Andrews Feb 7 '18 at 20:17
  • $\begingroup$ Thanks for the comment. My point was that you can not have convergence under the given condition at all. $\endgroup$ – Mohammad Riazi-Kermani Feb 7 '18 at 20:28

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