Determine whether the sum of 1/(1 + n (ln n)^2) from n = 1 to infinity converges, and estimate its value.

Step-by-step solution

1. 1

   Deal with the first term separately. At $n=1$, $\ln 1=0$, so the term is $\dfrac{1}{1+0}=1$. The interesting behaviour starts at $n=2$, so write

   $S=1+\sum_{n=2}^{\infty}\frac{1}{1+n\left(\ln n\right)^{2}}$

2. 2

   Compare with the Bertrand series. For $n\ge 2$ the $+1$ only makes the denominator bigger, so

   $0<\frac{1}{1+n\left(\ln n\right)^{2}}<\frac{1}{n\left(\ln n\right)^{2}}$

   and $\sum_{n\ge 2}\dfrac{1}{n(\ln n)^{p}}$ converges exactly when $p>1$. Here $p=2$, so the comparison series converges and the given series converges too.

3. 3

   See why $p>1$ is the threshold. The substitution $u=\ln x$ turns the integral test into

   $\int_{2}^{\infty}\frac{dx}{x(\ln x)^{2}}=\int_{\ln 2}^{\infty}\frac{du}{u^{2}}=\frac{1}{\ln 2}$

   finite because the exponent on $u$ exceeds 1. With $p=1$ the same substitution gives $\int du/u$, which diverges — the classic borderline case.

4. 4

   Accept that no elementary closed form exists. Unlike $\sum 1/n^{2}=\pi^{2}/6$, this sum has no known expression in terms of standard constants, so the honest answer is a convergence verdict plus a numerical value.

5. 5

   Estimate the value with a tail integral, not a raw partial sum. Convergence is extremely slow: the tail after $N$ terms is approximately

   $\int_{N}^{\infty}\frac{dx}{x(\ln x)^{2}}=\frac{1}{\ln N}$

   At $N=10^{7}$ the partial sum is $2.42985$ but the tail is still $0.06204$ — so stopping there and reporting $2.43$ (or $2.42$) understates the answer by more than $2\%$.

6. 6

   Combine partial sum and tail. Adding the two pieces at four different cut-offs gives a stable value:

   $N=10^{4}:\ 2.38332+0.10857=2.49189,\qquad N=10^{7}:\ 2.42985+0.06204=2.49189$

   so

   $\sum_{n=1}^{\infty}\frac{1}{1+n\left(\ln n\right)^{2}}\approx 2.4919$

   The agreement across three orders of magnitude in $N$ is what makes the estimate trustworthy.