Let r<p. Prove that if the series sum from n=1
Let $r<p$. Prove that if the series $\displaystyle\sum_{n=1}^{\infty}\left(\sum_{v=1}^{n}|u_v|^r\right)^{1/r}$ converges, then the series $\displaystyle\sum_{n=1}^{\infty}\left(\sum_{v=1}^{n}|u_v|^p\right)^{1/p}$ also converges.
Explanation of Method
The key fact is that for any fixed finite vector, its unnormalized $\ell^s$-norm $\left(\sum|a_i|^s\right)^{1/s}$ is non-increasing as the exponent $s$ increases. Applying this to the partial vector $(u_1,\ldots,u_n)$ with $s=r$ versus $s=p>r$ shows $S_n^{(p)}\le S_n^{(r)}$ term by term, and the comparison test does the rest.
- Prove the norm-monotonicity lemma: for $r<p$, $\|a\|_p\le\|a\|_r$ for any finite or infinite sequence.
- Apply it to each partial vector $(u_1,\ldots,u_n)$.
- Compare term by term and conclude via the comparison test.
Step 1 · Norm-monotonicity lemma
Let $M=\left(\sum_i|a_i|^r\right)^{1/r}>0$ be finite. Since $|a_i|^r\le\sum_k|a_k|^r=M^r$, each $|a_i|\le M$, so $|a_i|/M\le1$. For $p>r$: $(|a_i|/M)^p\le(|a_i|/M)^r$ (raising a number in $[0,1]$ to a larger power only shrinks it). Summing: $\sum_i(|a_i|/M)^p\le\sum_i(|a_i|/M)^r=1$, so $\sum_i|a_i|^p\le M^p$, i.e. $\|a\|_p\le M=\|a\|_r$.
Step 2 · Apply to each partial vector
Fix $n$ and apply Step 1 to $(u_1,\ldots,u_n)$: $$ S_n^{(p)}=\left(\sum_{v=1}^n|u_v|^p\right)^{1/p} \le \left(\sum_{v=1}^n|u_v|^r\right)^{1/r} = S_n^{(r)}. $$
Step 3 · Compare and conclude
Since $0\le S_n^{(p)}\le S_n^{(r)}$ for every $n$, and $\sum_{n=1}^{\infty}S_n^{(r)}$ converges by hypothesis, the direct comparison test gives $\sum_{n=1}^{\infty}S_n^{(p)}$ converges too.
Summary Table
| Quantity | Relation |
|---|---|
| $S_n^{(p)}$ vs $S_n^{(r)}$ | $S_n^{(p)}\le S_n^{(r)}$ for $r<p$ |
| Conclusion | $\sum S_n^{(p)}$ converges by comparison |
Additional Notes
- The lemma in Step 1 is the discrete version of the classical fact $\ell^r\subset\ell^p$ for $r<p$: any sequence with finite $r$-th power sum automatically has finite $p$-th power sum, with the norm only shrinking as $p$ grows.
- The proof works verbatim whether the inner sum runs over a finite range (as here, $v=1,\ldots,n$) or is genuinely infinite, since the normalization trick $|a_i|/M\le1$ never uses finiteness of the index set beyond $M<\infty$.
- This lemma is the same tool underlying Problem 152's double-series result, where it is applied to each "row" of an infinite matrix rather than to a growing partial vector.
Result
The implication holds via direct comparison: