# $\sum_{n=3}^\infty \frac{1}{n(\ln n)^4}$ what upper bound does it yield for the error S-S30

The integral test enables us to bound the error approximation of the series $$S=\sum_{n=3}^\infty \frac{1}{n(\ln n)^4}$$

by the partial sum $$S30=\sum_{n=3}^{30} \frac{1}{n(\ln n)^4}$$

What upper bound does it yield for the error S−S30 ? Give your answer accurate to 3 signiﬁcant digits.

SO I calculate $$\int_{30}^ \infty \frac{dn}{n(\ln n)^4}$$ but the answer is wrong why appreciate any help

• What did you get as your answer? – Simply Beautiful Art Mar 3 '17 at 17:06

You've included the $n=30$ term in your partial sum, so you don't want to count it again in your error term.
$$\int_N^\infty f(x)\, dx \leq \sum_{n=N}^\infty f(n)\leq f(N)+\int_N^\infty f(x)\,dx$$
Letting $f(n)=\frac{1}{n(\ln n)^4}$, the error between the partial sum $S_{30}=\sum_{n=3}^{30}f(n)$ and the exact value infinite sum $S=\sum_{n=3}^\infty f(n)$ is the tail of the series $S=\sum_{n=31}^\infty f(n).$ So this is the sum we want an upper bound for, which the integral test tells us is $f(31)+\int_{31}^\infty f(x)\,dx$.