# Limiting behaviour of \$sum_{j=0}^infty|sum_{k=1}^na_{j+k}|^p\$ Code Answer

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I would like to prove or disprove the following statement.

Suppose that \${a_j:jge0}\$ is an absolutely summable real sequence, i.e. \$sum_{j=0}^infty|a_j|<infty\$. Then
\$\$
\$\$
where \$1<p<2\$.

For example, if \$a_j=r^j\$ with \$|r|<1\$, then the statement is true and even stronger result holds since
\$\$
\$\$

I tried to use Hölder’s inequality and the triangle inequality for the \$ell^p\$ norm, but I could not obtain the result. By Hölder’s inequality,
\$\$
sum_{j=0}^inftybiggl|sum_{k=1}^na_{j+k}biggr|^p
lesum_{j=0}^inftybiggl|Bigl(sum_{k=1}^n|a_{k+j}|^pBigr)^{1/p}n^{1-1/p}biggr|^p=n^{p-1}sum_{k=1}^nsum_{j=0}^infty|a_{k+j}|^ple n^psum_{j=0}^infty|a_j|^p,
\$\$
but this inequality is not accurate enough. The assumption of absolute summability needs to be somehow used, but I have no idea how to do that.

Any help is much appreciated!

Let \$s = sum_{jge 0} |a_{j}|\$ and \$S_{j} = a_{j+1}+ldots+a_{j+n}\$. As \$s<infty\$ one has that there exists \$Nge 0\$ such that \$sum_{jge N}|a_j|<1\$. Therefore for any \$jge N\$ any \$nin {mathbb N}\$, it can bee seen that
\$\$
|S_j| = |a_{j+1}+ldots+a_{j+n}|le |a_{j+1}|+ldots+|a_{j+n}|le sum_{jge N}|a_j| <1.
\$\$
Thus as \$pge 1\$, one has \$|S_j|^p<|S_j|\$ and consequently
\$\$
|S_j|^ple |S_j|le |a_{j+1}|+ldots+|a_{j+n}|
\$\$
So
\$\$
sum_{jge N} |S_j|^p
le
sum_{j>N}|S_j|le sum_{jge N} |a_{j+1}|+ldots+|a_{j+n}|= sum_{jge N} |a_{j+1}|+ldots+sum_{jge N}|a_{j+n}|
\$\$
As for any \$k=2,ldots,n\$ one has that
\$\$
sum_{jge N} |a_{j+k}|le sum_{Nle j le N+k-2} |a_{j+1}| + sum_{jge N} |a_{j+k}|= sum_{jge N} |a_{j+1}|,
\$\$
it can be seen that
\$\$
sum_{jge N} |S_j|^ple n sum_{jge N} |a_{j+1}|
\$\$
Therefore
\$\$
sum_{jge 0} |S_j|^p< n sum_{jge N} |a_{j+1}| + sum_{ 0le j< N} |S_j|^p.
\$\$
So as \$sum_{jge N} |a_{j+1}|<1\$
\$\$
sum_{jge 0} |S_j|^p< n + c.
\$\$
where \$c = sum_{ 0le j< N} |S_j|^p\$ is a constant.