Approximate the sum of the series (Sigma) 1/(n^3) by using the first 5 terms.

Which is just: 1+1/8+1/27+1/64+1/125 = 1.18543....

How do I estimate the error involved in this approximation?

Question

Approximate the sum of the series (Sigma) 1/(n^3) by using the first 5 terms.

Which is just: 1+1/8+1/27+1/64+1/125 = 1.18543....

How do I estimate the error involved in this approximation?

amistre64 · Answer

the eror is never greater than the next term, or something along those lines

anonymous · Answer

So I would be comparing the error of 1.18543 and 1.18543 + 1/(6^3)?

amistre64 · Answer

wish i could tell, i never really got that far with it :)

amistre64 · Answer

http://ecademy.agnesscott.edu/~lriddle/apcalculus/approxSeries.pdf might be useful

amistre64 · Answer

total sum = partial + remainder
$$s=s_n+R_n$$
$$R_n=s-s_n$$

$$R_{max}=\int_{n+1}^{inf} f(x)dx$$
$$R_{min}=\int_{n}^{inf} f(x)dx$$

amistre64 · Answer

i think i might have those backwards tho

amistre64 · Answer

anyhoos

$$\int_{6}^{inf}x^{-3}dx=-\frac{1}{2(inf)^2}+\frac{1}{2(6)^2}$$
$$\int_{5}^{inf}x^{-3}dx=-\frac{1}{2(inf)^2}+\frac{1}{2(5)^2}$$

so it looks like the total sum rests between sn + 1/72 and sn+1/50

amistre64 · Answer

well, sn being the 5th partial sum in this case