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I have $\sum(\frac{(-1)^n}{\sqrt{n}} + \frac1n)$.

Nth test: $\lim_{n->\infty}{(\frac{(-1)^n}{\sqrt{n}} + \frac1n)}$ = 0.

I think that we can not split it into two sums like $\sum(\frac{(-1)^n}{\sqrt{n}}) + \sum(\frac1n)$ because the second one is divergence.

I know that sum is absolutely divergence but how can I prove relative divergence? All tests say about the sum convergence but not divergence.

jimjim
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IlFiltsin
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  • You can split into two different sums since addition is an associative operation, i.e (a+b)+c=a+(b+c). Then since we know the second series diverges, then the original series must diverge too. – Alessio K Dec 26 '19 at 23:07
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    If $\sum b_n$ diverges and $\sum a_n$ converges, can $\sum (a_n+b_n)$ converge? HINT: $b_n = (a_n+b_n) + (-a_n)$. – Ted Shifrin Dec 26 '19 at 23:07
  • I think it is can not. But however, I need to find some math proof. Thank you for your hint but I can't come up with it. – IlFiltsin Dec 26 '19 at 23:29
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    I told you how to do the proof. :) It is an argument by contradiction. Suppose $\sum (a_n+b_n)$ does converge. Then ... – Ted Shifrin Dec 26 '19 at 23:40

1 Answers1

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The pertinent theorem is that if $\sum a_n$ and $\sum b_n$ are both convergent, then so is $\sum(a_n+b_n)$. This theorem doesn't care whether the convergence is conditional or absolute.

So if $\sum\left({(-1)^n\over\sqrt n}+{1\over n}\right)$ were convergent, then, since $-\sum{(-1)^n\over\sqrt n}$ is convergent, we would have to conclude that $\sum\left({(-1)^n\over\sqrt n}+{1\over n}-{(-1)^n\over\sqrt n}\right)=\sum{1\over n}$ is convergent, which we know not to be the case.

The key to proving the pertinent theorem is to note that if $\left|\sum_{n=1}^Na_n-A\right|\lt\epsilon/2$ and $\left|\sum_{n=1}^Nb_n-B\right|\lt\epsilon/2$, then

$$\left|\sum_{n=1}^N(a_n+b_n)-(A+B)\right|\le\left|\sum_{n=1}^Na_n-A\right|+\left|\sum_{n=1}^Nb_n-B\right|\lt\epsilon/2+\epsilon/2=\epsilon$$

Barry Cipra
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