Estimating the Value of a Series — Question 7

*Question 7 For $\displaystyle S=\sum_{n=2}^{\infty}\frac1{n(\log n)^2}$, derive a two-sided Integral Test estimate for $R_N=S-S_N$. Then determine what the upper bound requires to guarantee $R_N<0.01$ and interpret the result.

*Question 7 – Solution

The function $f(x)=1/[x(\log x)^2]$ is positive and decreasing for $x\ge2$.

With $u=\log x$, $$\int_A^{\infty}\frac{dx}{x(\log x)^2}=\int_{\log A}^{\infty}u^{-2}\,du=\frac1{\log A}.$$ Therefore $$\boxed{\frac1{\log(N+1)}\le R_N\le\frac1{\log N}}.$$

$$\frac1{\log N}<0.01\iff \log N>100\iff N>e^{100}.$$ Thus the least certified integer is $\lfloor e^{100}\rfloor+1$, an astronomically large index (about $2.69\times10^{43}$).

The lower and upper bounds are asymptotically equal, so this is not merely a loose estimate: the series itself converges extraordinarily slowly.