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

Question

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

Answer 1

You've included the $n=30$ term in your partial sum, so you don't want to count it again in your error term.

In more detail, to start the standard form of the integral test takes the form

$$ \int_N^\infty f(x)\, dx \leq \sum_{n=N}^\infty f(n)\leq f(N)+\int_N^\infty f(x)\,dx $$

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 $f(31)+\int_{31}^\infty f(x)\,dx$.