summation - Transformed Sum question

I am trying to understand how this sum was transformed from

$$\sum_{n=1}^\infty \frac {\sqrt{n}}{n(n+1)}$$

to

$$1 + \sum_{n=2}^\infty \frac{\sqrt{n}-\sqrt{n-1}}{n}$$

I see that the index was changed from $n=1$ to $n=2$, thus requiring that the case for $n=1$ be added but I get $\frac{1}{2}$. Not sure where the $1$ comes from and how they transformed the rest of the sum.

Answer

Note that

$$\begin{align} \sum_{n=1}^N\left(\frac{\sqrt n}{n(n+1)}\right)&=\sum_{n=1}^N\left(\frac{\sqrt{n}}n-\frac{\sqrt n}{n+1}\right)\\\\ &=\color{blue}{\sum_{n=1}^N\left(\frac{\sqrt{n}}n\right)}-\color{red}{\sum_{n=1}^N\left(\frac{\sqrt{n}}{n+1}\right)}\\\\ &=\color{blue}{1+\sum_{n=2}^N\left(\frac{\sqrt n}{n}\right)}-\color{red}{\sum_{n=1}^N\left(\frac{\sqrt{n}}{n+1}\right)}\\\\ &=\color{blue}{1+\sum_{n=2}^N\left(\frac{\sqrt n}{n}\right)}-\color{red}{\sum_{n=2}^N\left(\frac{\sqrt{n-1}}{n}\right)-\frac{\sqrt {N}}{N+1}}\\\\ &=1+\sum_{n=2}^N\left(\frac{\sqrt n-\sqrt{n-1}}{n}\right)-\frac{\sqrt{N}}{N+1}\\\\ \end{align}$$

Taking the limit as $N\to \infty$ shows that

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

as was to be shown!