Re: Truncation of infinite series

From: Karl (breitu_at_arcor.de)
Date: 10/08/04 

Date: Fri, 08 Oct 2004 09:01:03 +0200

> The ratio between successive terms of the series is r = (i+1/i)^(2/3) *
> x, so the absolute
> error of truncating at the n-th term, u_n, is
>
> E(n) <= (r * u_n ) / (1-r)
>
> How can I determine n to limit the relative error? There must be
> something simple I am
> missing here.
>
As Virgil wrote, the relative error is the absolute divided by the
correct value, also if S(x) denotes the value of the infinite sum:

Err_rel.=E(n)/S(x)

Since the sum of the series is strictly increasing, an upper bound is
if S(n,x) denotes the sum up to n:

Err_rel.<= E(n)/S(n,x)

If you want a simple formula, get a simple lower bound for S(n,x), for
example \sum_1^n x^i .

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