remainder estimate for integral test

[Guest]

Hello all - I was browsing the internet to help study for my upcoming calculus exam and came upon the following problem:


(a) Find the partial sum S[10] of the series sum(1/(n^4),n = 1 .. infinity) . Estimate the error in using S[10] as an approximation to the sum of the series.


Does anyone know how to begin this? My teacher briefly went over this topic however I am still struggling with it.

Any kind of hints would be greatly appreciated.

Thank you

 

[galactus]

There is a general formula for series of this form, believe it or not.

\(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\frac{(-1)^{k-1}B_{2k}(2{\pi})^{2k}}{2(2k)!}\)

\(\displaystyle B_{2k}\) is a Bernouilli number, which I won't get into.

For your particular series:

\(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{n^{2(2)}}=\frac{(-1)^{2-1}B_{2(2)}(2{\pi})^{2(2)}}{2(2(2)!)}\)

The 4th Bernoulli number is -1/30.

So, \(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{(-1)(\frac{-1}{30})(2{\pi})^{4}}{48}\)

=\(\displaystyle \H\\\frac{{\pi}^{4}}{90}\)

That is the sum of your series from 1 to infinity.

You can calculate S(10) and find the difference.