Integral test error approximation Find an N so that:
$\sum_{n=1}^\infty \frac{1}{n^4}$ is between $\sum_{n=1}^N \frac{1}{n^4}$ and $\sum_{n=1}^N \frac{1}{n^4} + 0.005$
I am getting N = 200.
Is this correct? I don't know if I am doing it correctly.
 A: You want $\sum_{n = N+1}^{\infty} \frac{1}{n^4} < .005$. 
We have the integral comparison 
$$\int_{N+1}^{\infty} \frac{1}{x^4}< \sum_{n = N+1}^{\infty} \frac{1}{n^4}< \int_{N}^{\infty} \frac{1}{x^4}$$
or 
$$\frac{1}{3(N+1)^3} < \sum_{n = N+1}^{\infty} \frac{1}{n^4} < \frac{1}{3N^3}$$
The equation $\frac{1}{3N^3} = 0.005$ has the solution $N = 4.05...$ and we see that 
$$\frac{1}{3 \cdot 5^3} < \sum_{n = 5}^{\infty} \frac{1}{n^4} < \frac{1}{3\cdot 4^3}$$
or 
$$ 0.00266 < \sum_{n = 5}^{\infty} \frac{1}{n^4} < 0.00520$$
Still that does not guarantee that the remainder starting with the fifth term is less than $0.005$ but we are close. Let's get a better approximation: 
$$\sum_{n = 5}^{\infty} \frac{1}{n^4} = \frac{1}{5^4} + \sum_{n = 6}^{\infty} \frac{1}{n^4}< \frac{1}{5^4}  + \frac{1}{3\cdot 5^3} = 0.0016+ 0.00266... = 0.00426... < 0.005$$
Thus the first such $N$ is $4$ ( anything larger OK too).
$\bf{Added:}$
Thanks to the observation of @Claude Leibovici:  we check
$$\frac{\pi^4}{90} - ( 1+ \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4}) =0.00357...$$ while
$$\frac{\pi^4}{90} - ( 1+ \frac{1}{2^4} + \frac{1}{3^4} ) = 0.00747...$$
