Question

find a value of n such that sn is within 0.0004

\[\sum_{n=1}^{\infty} \frac{2}{n^{5}}\]

Answer 1

this reminds me of a remainder function thingy

Answer 2

\[S=S_n+R_n\]\[R_n < .0004\]if memory serves

Answer 3

\[R_n: 2\int_n^{\infty}x^{-5}dx<.0004\]
\[\int_n^{\infty}x^{-5}dx<.0002\]
\[-\frac{1}{4}\infty ^{-4}+\frac{1}{4}n^{-4}<.0002\]
\[\frac{1}{4}n^{-4}<.0002\]
\[n^{-4}<.0008\]
\[n^{4}>\frac{1}{.0008}\]
\[n>\sqrt[4]{\frac{1}{.0008}}\]
maybe

Answer 4

n > 5.95

so id chk at n=5 and n=6 to be safe

Answer 5

to inf = x.0738555...
to 5 = x.0733235...
to 6 = x.0735807...

.0738555
.0735807
----------
.0002748

.0738555...
.0733235...
-----------
.0005320

id say n=6 wins

Answer 6

how did you get
the values:

to inf = x.0738555...
to 5 = x.0733235...
to 6 = x.0735807...

Answer 7

i plugged the summation into wolframalpha.com to make the dbl chk easier for me

Answer 8

oh I see thanks I must have been typing in my problem incorrectly

Answer 9

sum (from a to b) f(n) i think is what i used

Answer 10

I used
https://www.wolframalpha.com/input/?i=sum+n+%3D+1+to+infinity+2%2Fn^%285%29

Answer 11

so you are looking for the value of n that is closest to 0.0004?

Answer 12

you are looking for the value of n that insures that you are no more than .0004 in error

Answer 13

the smallest n that is

Answer 14

(sum at infinity) - (sum at n) is less than .0004

Answer 15

since:

\[S_{\infty}=S_n+R_n \]then\[S_{\infty}-S_n=R_n \]

but we want it such that

\[S_{\infty}-S_n<.0004\]which means that we need to find a suitable
\[R_n < .0004\]

Answer 16

the remainder part is going to fall someplace between \(\int R_n\) and \(\int R_{n+1}\)

Answer 17

\[\int_{n}^{\infty}f(x)dx\ \le \sum_{n}^{\infty}f(n)\le\ \int_{n+1}^{\infty}f(x)dx\]

or something like that

Answer 18

i prolly got my n and n+1 mixed around tho

Answer 19

I see thanks