The upper bound of the sum of a series

Question

If $y_n$ is a nonincreasing series of real numbers and $0\leq y_n\leq1 $ for $n\geq 0$, $y_0 = 1$ and we know that $\sum_{n=0}^\infty y_n \leq A $, then is there a way to find a tight upper bound for $\sum_{n=0}^\infty y_n^m $, where $m$ is an integer number?

Answer 1

For positive integers $m$, let $N$ be an index for which $y(n)<1$ for all $n>N$. An upper bound for $\sum_{n=0}^{\infty}(y(n))^m$ is $$\sum_{n=0}^{N}(y(n))^m+A-\sum_{n=0}^{N}y(n)$$

Answer 2

A different bound: $$ \sum_{n=0}^\infty y_n^m=y_0^m\sum_{n=0}^\infty\Bigl(\frac{y_n}{y_0}\Bigr)^m\le y_0^m\sum_{n=0}^\infty\frac{y_n}{y_0}=y_0^{m-1}\,A. $$

Answer 3

Assuming positive terms,

$$y_n\ge1\implies y_n^m\ge y_n$$ then as the original series converges

$$B:=\sum_{n=0,y_n\ge1}^\infty(y_n^m-y_n)$$

is finite and

$$\sum_{n=0}^\infty(y_n^m-y_n)\le B$$ or $$\sum_{n=0}^\infty y_n^m<A+B.$$