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If I have an infinite series $a_n$ such as $\frac{1 + 4^n}{1 + 3^n}$ and $a_n$ is equal to the infinite series $b_n + c_n$ or $\frac{1}{3^n} + \frac{4^n}{3^n}$, is it possible to show in a question that the series $a(n)$ diverges by showing that the series $c(n)$ or $\frac{4^n}{3^n}$ diverges? I was doing a question and this is how I showed it. But when I checked the answer in my textbook they proved it by comparing the series $a_n$ to the different series $\frac{1 + 4^n}{3^n}$ and found the limit of the ratio of those series. That method also makes sense to me too but I find my way easier but I don't know if my way is allowed? My way makes sense to me because if I have a series and it's equal to $a + b$ of course both a and b have to converge for the whole series to converse. You can't have part of the series converge and the other part diverge. Could someone clear my confusion. Thank You.

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One common sense rule (which you intuited): $\sum a_n$ diverges $\implies \sum a_n+b_n$ diverges provided that $a_n,b_n$ are of the same parity $\forall n$. This is far from necessary or sufficient conditions, but it accords with your intuitive idea. Note however that if the $a_n,b_n$ are of different parity, then even though one of the sequences may diverge on its own, the sum may "cancel" in some way and divergence will not be guaranteed. – Coffee_Table Mar 24 '13 at 3:32
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Look for the various series (and sequence) convergence criteria, check how they are proved. To see why they are just that way, poke around looking at how you can make the convergence fail by violating their conditions. Yes, many of those are quite loose tests, in that you can bend them quite out of shape in just the right way and the series continues converging, but capturing all the variants just would give a terrible mess. Look for divergence criteria too, apply the same third degree.

You should apply this same idea to other results you get thaught.

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