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# series and intergrals watch

1. Hi im stuck on the last part of a question in the first part you worked out that

the sum between (M+1) and N of 1/(r^3) is

less than intergral between m and n of 1/(x^3) and greater than the integral between (m+1) and ( N+1) of 1/(x^3)

then the next part which i am stuck on goes

given that the sum between 1 and 25 of 1/(r^3) =1.20129 to 5 dp and that that the series sum between 1 and infinity of 1/(r^3) is convergent show that

1.2020<sum between 1 and infinity 1/(r^3) <1.2021

thanks alot
2. Using the first part,

(int from r = 26 to infinity) 1/r^3 dr
<= (sum from r = 26 to infinity) 1/r^3
<= (int from r = 25 to infinity) 1/r^3 dr

You can't say what (sum from r = 26 to infinity) 1/r^3 equals, but you can find two numbers k and K such that

k <= (sum from r = 26 to infinity) 1/r^3 <= K

--

Once you have k and K, think about

(sum from r = 1 to infinity) 1/r^3
= (sum from r = 1 to 25) 1/r^3 + (sum from r = 26 to infinity) 1/r^3

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