$\sum_{n=1}^{\infty} \frac{n^2}{(1+n^3)^p}$, find values of p I have a sum $\sum_{n=1}^{\infty} \frac{n^2}{(1+n^3)^p}$ and $p>1$. Use integral tests and determine which values of $p$ can the partial sum, $s_5$ estimates true sum of $s$ to within $10^{-6}$ error. So this is a question about errors. It says using $s_5$, so error formula is $error \leq \int_{5}^{\infty} \frac{n^2}{(1+n^3)^p}$. So I calculate the sum and I can get p?  Here I need $10^{-6}$ maximum error, so I equal
$10^{-6} \leq \int_{5}^{\infty} \frac{n^2}{(1+n^3)^p}$. And finally it's get $p = 3.246$.
 A: You have
$$S_p=\sum_{n=1}^{\infty} \frac{n^2}{(1+n^3)^p}\qquad \text{and} \qquad I_p=\int_{5}^{\infty} \frac{n^2}{(1+n^3)^p}\,dn$$The integral is easy to compute since
$$J=\int \frac{n^2}{(1+n^3)^p}\,dn=\frac 13\int \frac{3n^2}{(1+n^3)^p}\,dn=\frac 13\int \frac{dx}{(1+x)^p}\,dx=\frac 13\int \frac{dy}{y^p}$$that is to say
$$I_p=\frac{3^{1-2 p} \,14^{1-p}}{p-1}$$ So, making the problem more general,you want to know $p$ such that
$$10^{-k} \leq \frac{3^{1-2 p} \,14^{1-p}}{p-1}$$ or
$$(p-1)\,126^{p-1} \geq 3 \times 10^k$$ Rewrite it as
$$\big[(p-1)\log(126)\big] e^{\big[(p-1)\log(126)\big]} \geq 3 \log(126)\times 10^k$$ and the solution of the equation is given is terms of Lambert function
$$(p-1)\log(126)=W\big[ 3 \log(126)\times 10^k \big]\implies $$ $$\color{blue}{p=1+\frac{W\big[ 3 \log(126)\times 10^k \big] }{\log(126)}}$$ Since that argument is large, you can evaluate it using
$$W(t)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(t)$ and $L_2=\log(L_1)$.
For $k=6$, this will give $p\sim 3.866$.
