Question

nth Partial Sum Find the first four partial sumsand the nth partial sum of the sequence $a_n$.

$a_n=\sqrt{n}-\sqrt{n+1}$.

Short answer

The partial sums are:

First partial sum is = $1-\sqrt{2}$.

Second partial sum is = $1-\sqrt{3}$.

Third partial sum is = $-1$.

Fourth partial sum is = $1-\sqrt{5}$.

nth partial sum is $\left(1-\sqrt{n}\right)$.

Step-by-step solution

Step 1.  Given

The given sequence is $a_n=\sqrt{n}-\sqrt{n+1}$.

Step 2.  To determine

We have to find thefirst four partial sums and the nth partial sum.

Step 3.  Calculation

The given sequence is $a_n=\sqrt{n}-\sqrt{n+1}$.

First partial sum is:

$\begin{array}{l}=\sqrt{1}-\sqrt{1+1}\\ =\sqrt{1}-\sqrt{2}\\ =1-\sqrt{2}\end{array}$

Second partial sum is:

$\begin{array}{l}=\left(\sqrt{1}-\sqrt{1+1}\right)+\left(\sqrt{2}-\sqrt{2+1}\right)\\ =\left(\sqrt{1}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{3}\right)\\ =1-\sqrt{3}\end{array}$

Third partial sum is:

$\begin{array}{l}=\left(\sqrt{1}-\sqrt{1+1}\right)+\left(\sqrt{2}-\sqrt{2+1}\right)+\left(\sqrt{3}-\sqrt{3+1}\right)\\ =\left(\sqrt{1}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{4}\right)\\ =1-\sqrt{4}\\ =1-2\\ =-1\end{array}$

Fourth partial sum is:

$\begin{array}{l}=\left(\sqrt{1}-\sqrt{1+1}\right)+\left(\sqrt{2}-\sqrt{2+1}\right)+\left(\sqrt{3}-\sqrt{3+1}\right)+\left(\sqrt{4}-\sqrt{4+1}\right)\\ =\left(\sqrt{1}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{4}\right)+\left(\sqrt{4}-\sqrt{5}\right)\\ =1-\sqrt{5}\end{array}$

From the sequence of the partial sums:

$\begin{array}{l}1-\sqrt{2},1-\sqrt{3},-1,1-\sqrt{5},\mathrm{....}\\ =\left(1-\sqrt{2}\right),\left(1-\sqrt{3}\right),\left(1-\sqrt{4}\right),\left(1-\sqrt{5}\right),\mathrm{....}\end{array}$

It is easy to see that the nth partial sum will be role="math" $\left(1-\sqrt{n}\right)$.

Hence, nth partial sum is $\left(1-\sqrt{n}\right)$.