Find an upper bound for the sum of 1/(1 + n^2 ln n) from n = 1 to infinity.

Step-by-step solution

1. 1

   Handle $n=1$ separately. Since $\ln 1=0$, the first term is $\dfrac{1}{1+0}=1$ — the largest term in the whole series, and one that no comparison involving $\ln n$ can cover. Split it off:

   $S=1+\sum_{n=2}^{\infty}\frac{1}{1+n^{2}\ln n}$

2. 2

   Drop the $+1$ to enlarge each term. For $n\ge 2$, $1+n^{2}\ln n>n^{2}\ln n>0$, so making the denominator smaller makes the fraction larger:

   $\frac{1}{1+n^{2}\ln n}<\frac{1}{n^{2}\ln n}$

   Every inequality in a bounding argument must push in the same direction; here both do.

3. 3

   Replace $\ln n$ by its smallest value on the range. For $n\ge 2$ we have $\ln n\ge\ln 2$, so

   $\frac{1}{n^{2}\ln n}\le\frac{1}{n^{2}\ln 2}$

   This is the step that converts an awkward log-weighted series into a plain $p$-series, at the cost of some tightness.

4. 4

   Sum the comparison series exactly. Using the Basel result minus its first term,

   $\sum_{n=2}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}-1\approx 0.644934$

5. 5

   Assemble the bound.

   $S<1+\frac{1}{\ln 2}\left(\frac{\pi^{2}}{6}-1\right)=1+\frac{0.644934}{0.693147}=1.93044$

   The series converges, and its sum is certainly below $1.93044$.

6. 6

   Judge how tight the bound is. Summing three million terms gives $S\approx 1.4976$, so the bound overshoots by about $0.43$. Most of the slack comes from step 3: replacing $\ln n$ by $\ln 2$ is generous for large $n$, where $\ln n$ is much bigger. Splitting the sum at, say, $n=10$ and using $\ln 10$ beyond that would tighten it considerably.