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Let $a_n$ be a sequence of positive reals such that $\sum_1^\infty a_n$ is finite then show that $\sum_1^\infty a_n^{1-\frac{1}{n}}$ is also finite.

N.B Is this statement true? I tried to use the limit comparison test but it's inconclusive. The converse is obviously true which makes it an "iff" statement. No well-known test seems to be working. Any help is appreciated.

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  • $\begingroup$ Could you give us some context here? Where did you encounter this statement? Do you have a reason to believe that it is true? $\endgroup$ – Omnomnomnom Aug 22 '18 at 17:03
  • $\begingroup$ particularly I don't have a valid reason I just tried to see by increasing $a_n$ a little bit whether the sum again converges. A few convergent series I have taken and found that the later converges on them and if the former diverges then the later also diverges $\endgroup$ – mudok Aug 22 '18 at 17:07
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    $\begingroup$ An observation: note that $$ \lim_{n \to \infty} \frac{a_n}{a_n^{1 - 1/n}} = \lim_{n \to \infty} a_n^{1/n} $$ So, by the limit comparison test, $a_n$ could only potentially be a counterexample if $a_n^{1/n} \to 0$. $\endgroup$ – Omnomnomnom Aug 22 '18 at 17:10
  • $\begingroup$ I felt that but could not come up with a counter. $\endgroup$ – mudok Aug 22 '18 at 17:13
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First, we note that $$ \lim_{n \to \infty} \frac{a_n}{a_n^{1 - 1/n}} = \lim_{n \to \infty} a_n^{1/n} $$ So, by the limit comparison test, $a_n$ could only potentially be a counterexample if $a_n^{1/n} \to 0$.

Now, if $a_n^{1/n} \to 0$, then we can write $a_n^{1/n} = \xi(n)$, where $\xi(n) \to 0$. That is, $a_n = \xi(n)^{1/n}$. With that, we have $$ \lim_{n \to \infty} (a_n^{1 - 1/n})^{1/n} = \lim_{n \to \infty} (a_n^{1/n})^{1 - 1/n} = \lim_{n \to \infty} \xi(n)^{1 - 1/n} $$ Let $y = \lim_{n \to \infty} a_n^{1 - 1/n}$. We have $$ \begin{align} \log(y) &= \lim_{n \to \infty} \log\left[ \xi(n)^{1 - 1/n}\right] = \lim_{n \to \infty} (1 - 1/n) \log(\xi(n)) = \lim_{n \to \infty} \log(\xi(n)) \end{align} $$ That is, we must have $\log(y) = -\infty$, and hence $y = 0$. That is, our sequence satisfies $(a_n^{1 - 1/n})^{1/n} \to 0$, which means that $a_n^{1 - 1/n}$ converges by the root test.

Thus, we conclude that in all cases where $\sum a_n$ is convergent, $\sum a_n^{1 - 1/n}$ must also be convergent.


A shorter proof: in the second case, we have $$ \lim_{n \to \infty} (a_n^{1 - 1/n})^{1/n} = \lim_{n \to \infty} (a_n^{1/n})^{1 - 1/n} = \left[\lim_{n \to \infty} a_n^{1/n} \right]^{\lim_{n \to \infty} (1 - 1/n)} = 0^1 = 0 $$ Since $0^1$ is not an indeterminate form, this manipulation is valid. Again, conclude that $\sum a_n^{1 - 1/n}$ converges by the root test.

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    $\begingroup$ How does the proof work if $a_n^{1/n}$ is a sequence with $\liminf$ equal to 0 but $\limsup$ equal to 1, for example? $\endgroup$ – Daniel Schepler Aug 22 '18 at 17:38
  • $\begingroup$ @DanielSchepler Good point, didn't think about that... Of course we can extract subsequences of $a_n^{1/n}$ which converge to $0$ and $1$, but I'm not quite sure where to go from there $\endgroup$ – Omnomnomnom Aug 22 '18 at 19:51
  • $\begingroup$ @DanielSchepler perhaps there's a more robust version of the limit comparison test involving liminfs/limsups $\endgroup$ – Omnomnomnom Aug 22 '18 at 19:53

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