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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?

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    $\begingroup$ $y(n)=-1$. Then $\sum_n y(n)<0$. But $\sum (y(n))^2$ is unbounded. Do you want the numbers to be non-negative? $\endgroup$
    – Peyton
    Commented Jul 31, 2017 at 10:31
  • $\begingroup$ yes, actually they are positive numbers all between 0 and 1. $\endgroup$ Commented Aug 1, 2017 at 2:56
  • $\begingroup$ and also $y_0 = 1$. $\endgroup$ Commented Aug 1, 2017 at 3:17

3 Answers 3

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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 $$\sum_{n=0}^{N}(y(n))^m+A-\sum_{n=0}^{N}y(n)$$

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

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  • $\begingroup$ Thank you very much, actually the $y_n$s are between 0 and 1, $y_0 = 1$ and I need a tighter bound for $\sum_{n=0}^\infty y_n^m$ than A. $\endgroup$ Commented Aug 1, 2017 at 3:13
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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 $$\sum_{n=0}^\infty y_n^m<A+B.$$

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