Estimate the sum from n=1 to infinity of (2n+1)^(−9) correct to five decimal places.

Question

simoner · Answer

$$\sum_{n=1}^{\infty}(2n+1)^{-9}$$

amistre64 · Answer

when is (2n+1)^(-9) <= 000001

amistre64 · Answer

spose to be a decimal place in front:  .000001

simoner · Answer

$$\frac{ 1 }{ 16n ^{8} }\le.00001$$

amistre64 · Answer

1/(2n+1) <= 9rt(.000001)

1/9rt(.000001) <= 2n + 1

1/9rt(.000001) - 1<= 2n

1/(2*9rt(.000001)) - 1/2<= n

amistre64 · Answer

looks like when n >= 1.8

rational · Answer

it is a decreasing function so the tail is trapped between Upper sum and Lower sum 

$$\int\limits_{k+1}^{\infty} (2x+1)^{-9}dx ~~~\le~~~ \sum\limits_{n=k+1}^{\infty} (2n+1)^{-9} ~~~\le~~~ \int\limits_{k}^{\infty} (2x+1)^{-9}dx $$

amistre64 · Answer

theres a thrm that says, for some type of function, yada yada, the error is at most the value of the first neglected term.

since n=2 is the first neglected term, then when n=1 we are within 5 decies of accuracy