Find the sum from n = 1 to infinity of 1 divided by [sqrt(n+1)(n+2) + (n+1)sqrt(n+2)].

Step-by-step solution

1. 1

   Factor the denominator. Write $n+2=\sqrt{n+2}\cdot\sqrt{n+2}$ and $n+1=\sqrt{n+1}\cdot\sqrt{n+1}$. The denominator becomes $\sqrt{n+1}\sqrt{n+2}\left(\sqrt{n+2}+\sqrt{n+1}\right)$ — a form no amount of expanding would reveal.

2. 2

   Rationalise the remaining sum of roots. Multiply numerator and denominator by $\sqrt{n+2}-\sqrt{n+1}$; since $\left(\sqrt{n+2}+\sqrt{n+1}\right)\left(\sqrt{n+2}-\sqrt{n+1}\right)=1$, the term becomes $\frac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n+1}\sqrt{n+2}}$.

3. 3

   Split the fraction. Dividing each part of the numerator by the denominator gives $\frac{1}{\sqrt{n+1}}-\frac{1}{\sqrt{n+2}}$ — a clean telescoping difference.

4. 4

   Write the partial sum. $S_N=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{N+2}}$, because every interior term cancels with its neighbour.

5. 5

   Take the limit. As $N\to\infty$ the term $\frac{1}{\sqrt{N+2}}\to 0$, so the sum is $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\approx 0.7071$.

6. 6

   Numerical check. Summing $2\times10^{6}$ terms gives $0.7063997$, and the closed form predicts $\frac{1}{\sqrt2}-\frac{1}{\sqrt{2000002}}=0.7063997$ — an exact match, including the slow tail.