If $\sum a_{n}$ is a real convergent series then what can we say about the convergence of the series $\sum a_{n}^{2k+1}$ i.e. odd power? It is convergent for positive term series. But what about any infinite series $\sum a_{n}.$ I tried to find counterexample but did't got. Please help me. Thanks in advance.
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Hint: comparison test. – bartgol Apr 15 '16 at 15:45
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@bartgol is it convergent??? – neelkanth Apr 15 '16 at 15:45
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1See also http://math.stackexchange.com/questions/607293/show-a-convergent-series-sum-a-n-but-sum-a-np-is-not-convergent – Clement C. Apr 15 '16 at 16:15
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My bad. I implicitly assumed $a_n$ to be, if not positive, at least alternating, in which case $\sum a_n^{2k+1}$ converges. But those are only particular cases. – bartgol Apr 15 '16 at 19:06
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No, it is not necessarily convergent. In fact, any function $f$ such that $\sum_j f(a_j)$ converges whenever $\sum_j a_j$ converges must be linear in some neighbourhood of $0$. See e.g. this question and answer.
Robert Israel
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@neelkanth You do not have $\sum_n a_n$ converges implies $\sum_n a_n^2$ converges. So the square function is not linear in any neighborhood indeed, but it does not satisfy the assumption either -- there is no contradiction. – Clement C. Apr 15 '16 at 16:07
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@Robert Israel No $\sum a_{n}^{2}$ is convergent if $\sum a_{n}$ for $a_{n}>0$ – neelkanth Apr 15 '16 at 16:09
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Yes, but the result stated above says "whenever $\sum_n a_n$ converges." There cannot be any restriction (e.g., positivity) on $a_n$: again, it is not true that $\sum_n a_n^2$ converges for all $(a_n)_n$ such that $\sum_n a_n$ converges. It is only true under some further restriction on $(a_n)$, so the result above does not apply. – Clement C. Apr 15 '16 at 16:11
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You didn't actually state you needed one (although it would be nice, yes): the answer above indeed does not provide one explicitly, but answers your question. – Clement C. Apr 15 '16 at 16:13
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Take the sum $\sum\limits_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}} = (\sqrt{2}-1) \zeta(1/2)\approx -0.6$. It's square will not converge. I think for uneven exponents it should however be true but I am not sure – Jannick Apr 15 '16 at 16:16
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@ClementC. Ah, I have now read http://math.stackexchange.com/questions/96666/if-sum-1-inftya-n3-diverges-does-sum-1-inftya-n You are right this should be closed as a duplicate. The simple example is $\dots,\frac{1}{n^{1/3}},\frac{1}{n^{1/3}},-\frac{2}{n^{1/3}},\dots$ After cubing you get $\dots,\frac{1}{n},\frac{1}{n},-\frac{8}{n},\dots$ which diverges. – almagest Apr 15 '16 at 16:29
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