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I am not quite sure how to phrase this question.

I have a series $$\sum_{n=1}^{\infty} |a_n|$$

I also have another series, for a natural number $N \geq 1$ $$\sum_{k=0}^{\infty}A_{k} = \sum_{n=N}^{\infty} |a_n|$$

Now, I am asked to write $\sum_{n=1}^{\infty} |a_n|$ in terms of $\sum_{k=0}^{\infty}A_{k}$

Since $\sum_{k=0}^{\infty}A_{k} = \sum_{n=N}^{\infty} |a_n|$, then (assuming I am correct) $$\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{N} |a_n| + \sum_{k=0}^{\infty}A_{k}$$

Now the question asks me to show that one series converges if and only if the other series converges.

First thing I thought was the limit comparison test

If I do limit comparison test on $\sum_{n=1}^{\infty} |a_n|$ and $\sum_{k=0}^{\infty}A_{k}$, does the starting point matter?($n=1 \quad n=N$)

If not, this will end up as $$\lim_{n \rightarrow \infty} \frac{|a_{n}|}{|a_{n}|} = 1$$

And since the limit exists, both series will have the same result for convergence test.

My concern is that I am not sure the steps I took were legit.

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  • $\begingroup$ Can you show us both the series if you don't mind? I'm guessing it's the harmonic function. $\endgroup$ – Ishraaq Parvez Apr 4 at 13:44
  • $\begingroup$ The question did not specify the series. Just $\sum_{n \geq 1} |a_{n}|$ $\endgroup$ – Sirou Ewei Apr 4 at 13:46
  • $\begingroup$ The "other series" definition is weird: is there any relation between $\;N,\,k\;$ or $\;A_k\;$ ? Or in fact are we to understand that $\;N\;$ is a definite, CONSTANT, natural number? $\endgroup$ – DonAntonio Apr 4 at 13:47
  • $\begingroup$ $N$ is an arbitrary natural number and should remain as a constant. I agree with you that "other series" sounds confusing, but that's what the original question said $\endgroup$ – Sirou Ewei Apr 4 at 13:54
  • $\begingroup$ This is way too complicated. Being convergent in $\mathbb{R}$ means being finite. If one series is finite, check what happens in the equality you first derived. $\endgroup$ – Meowdog Apr 4 at 14:14
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We have in fact that

$$\sum_{n=1}^{\infty} |a_n| = \sum_{r=1}^{N-1} |a_r| + \sum_{k=0}^{\infty}A_{k}$$

or using limits of the corresponding seuqneces of partial limits:

$$\lim_{n\to\infty}\sum_{j=1}^n|a_j|=\sum_{r=1}^{N-1}|a_r|+\lim_{K\to\infty}\sum_{k=0}^KA_k$$

Observe that the first term in the right side is a finite number as it is a finite sum.

And since all the series (finite and infinite) are of positive terms, comparison and arithmetic of limits come in handy: if $\;\sum\limits_{k=0}^\infty A_k=T\;$ converges, then the sequence $\;\left\{\,\sum\limits_{j=1}^n|a_j|\,\right\}_{n=}^\infty\;$ is monotonic ascending and bounded above, by the number $\;\sum\limits_{r=1}^{N-1} |a_r|+A\;$ and it thus converges. and something similar the other way around: if $\;\sum\limits_{j=}^\infty |a_j|=S\;$ converges, then by arithmetic of limits

$$\lim_{K\to\infty}\sum_{k=0}^KA_k=\lim_{n\to\infty}\sum_{j=2}^n|a_j|-\sum_{r=1}^{N-1}|a_r|=S-\sum_{r=1}^{N-1}|a_r|$$

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