Limsup, showing that two expressions are equal

Question

I am stuck at this problem which I use for something else.

If $\{a_i\}$ is a sequence of number then I want to prove that

$\limsup |a_i|^{1/i}=\limsup|a_{i+k}|^{|1/i}$, where k is a fixed positive integer.

My idea is as follows. Obviously:

$\limsup |a_i|^{1/i}=\limsup|a_{i+k}|^{1/(k+i)}$, it is just a different way to write the same quantity.

Then I tried this trick:

$\limsup|a_{i+k}|^{1/(k+i)}=\limsup(|a_{i+k}|^{(1/i)})^{i/(i+k)}$. So the problem is now reduced to showing:

That if we have a sequence $\{B_n\}$, then:

$\limsup |B_n|=\limsup|B_n|^{i/(i+k)}$.

And since the exponent goes to 1 it is almost obvious. If I was working with only lims this would not create any problems, but with lim sup I am not sure.

Answer 1

$\limsup x_i$ is the infimum of numbers $M$ such that for every $\epsilon>0$ we have $x_i\le M+\epsilon$ for all large enough $i$. Thus, it suffices to show that the statements

1. for every $\epsilon>0$ we have $|a_i|^{1/i}\le M+\epsilon$ for all large enough $i$
2. for every $\epsilon>0$ we have $|a_i|^{1/(i+k)}\le M+\epsilon$ for all large enough $i$

are equivalent: if $M$ satisfies one, then it satisfies the other too. This becomes more evident if 2 is rewritten as

3. for every $\epsilon>0$ we have $|a_i|^{1/i}\le (M+\epsilon)^{(i+k)/i}$ for all large enough $i$

Informally, the effect of $(i+k)/i$ is drowned out by the fact that we can put $\epsilon/2$ or $2\epsilon$ in these inequalities.

Answer 2

As you have noted above, the original problem will follow from this result:

$\underline {Lemma}$ If $a_n\ge0$ for all $n$ and $\displaystyle\lim_{n\to\infty}d_n=1$, then $\limsup a_n=\limsup (a_n)^{d_n}$.

$\underline {Proof}$ Let $s=\limsup a_n$ and $t=\limsup (a_n)^{d_n}$, $\;$where possibly $s=\infty$ or $t=\infty$.

Then $\displaystyle\lim_{k\to\infty}a_{n_{k}}=s$ for some subsequence $(a_{n_{k}})$ of $(a_n)$,

so $\displaystyle\lim_{k\to\infty}\left(a_{n_{k}}\right)^{d_{n_k}}=s$ and therefore $s\le t$.

Since $\big(\left(a_{n}\right)^{d_{n}}\big)^{\frac{1}{d_n}}=a_n$, applying the same argument shows that $t\le s$.

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(This argument uses that the lim sup is the least upper bound of the subsequential limits, and is itself a subsequential limit.)