$\sum\limits_{n=2}^∞ = \frac{1}{n(ln(n)^4)}$
the error in the approximation s≈sn (where s is the value of the infinite sum and sn is the n-th partial sum) is|s−sn|≤ ?
My first misunderstanding is regarding the difference between the infinite sum and the sum to the n-th term is. Aren't these two values technically the same? If n is ∞ wouldn't the value of the infinite sum be the same as the value of the nth partial sum since the nth term is ∞?
Integral Test:
$\int\limits_{n=2}^∞ = \frac{1}{n(ln(n)^4)} dn$
Let u = ln(n) du = $\frac{1}{n}$
$\int\limits_{n=2}^∞ = \frac{1}{n(u^4)}dx$
$\int\limits_{n=2}^∞ = u^-4 (du)$
$\int\limits_{n=2}^∞ = -\frac{1}{3(u^3)}$
$\int\limits_{n=2}^∞ = -\frac{1}{3(ln(n)^3)}$
$\lim\limits_{n->∞} = -\frac{1}{3(ln(n)^3)}$
=$\lim\limits_{n->∞} = -\frac{1}{3(ln(2)^3)}$
This value is the infinite sum I believe. If I subtract from it the partial sum then it would be 0.
The series $\sum_{n\geq 2}\frac{1}{n\log^4 n}$ is convergent (by Cauchy condensation test) but it converges very slowly. The function $f(x)=\frac{1}{x\log^4 x}$ is continuous, positive and decreasing on $(2,+\infty)$, hence
$$ \sum_{n\geq N}\frac{1}{n\log^4 n}\leq \int_{N-1}^{+\infty}\frac{dx}{x\log^4 x}=\frac{1}{3\log^3(N-1)} $$ and $$ \sum_{n\geq N}\frac{1}{n\log^4 n}\geq \int_{N}^{+\infty}\frac{dx}{x\log^4 x}=\frac{1}{3\log^3(N)}. $$