Question

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

$a_n=\log \left(\frac{n}{n+1}\right)$. [Hint: Use a property of logarithms to write the nth term as a difference.]

Short answer

The partial sums are:

First partial sum is = $-\log \left(2\right)$.

Second partial sum is = $-\log \left(3\right)$.

Third partial sum is = role="math" localid="1648590381032" $-\log \left(4\right)$.

Fourth partial sum is = $-\log \left(5\right)$.

nth partial sum is $-\log \left(n+1\right)$.

Step-by-step solution

Step 1.  Given

The given sequence is $a_n=\log \left(\frac{n}{n+1}\right)$.

Step 2. To determine

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

Step 3.  Calculation

The given sequence is $a_n=\log \left(\frac{n}{n+1}\right)$.

$a_n=\log \left(n\right)-\log \left(n+1\right)$[Using the quotient property of logarithm].

First partial sum is:

$\begin{array}{l}=\log \left(1\right)-\log \left(1+1\right)\\ =\log \left(1\right)-\log \left(2\right)\\ =0-\log \left(2\right)\\ =-\log \left(2\right)\end{array}$

Second partial sum is:

$\begin{array}{l}=\left[\log \left(1\right)-\log \left(1+1\right)\right]+\left[\log \left(2\right)-\log \left(2+1\right)\right]\\ =\left[\log \left(1\right)-\log \left(2\right)\right]+\left[\log \left(2\right)-\log \left(3\right)\right]\\ =\log \left(1\right)-\log \left(3\right)\\ =0-\log \left(3\right)\\ =-\log \left(3\right)\end{array}$

Third partial sum is:

$\begin{array}{l}=\left[\log \left(1\right)-\log \left(1+1\right)\right]+\left[\log \left(2\right)-\log \left(2+1\right)\right]+\left[\log \left(3\right)-\log \left(3+1\right)\right]\\ =\left[\log \left(1\right)-\log \left(2\right)\right]+\left[\log \left(2\right)-\log \left(3\right)\right]+\left[\log \left(3\right)-\log \left(4\right)\right]\\ =\log \left(1\right)-\log \left(4\right)\\ =0-\log \left(4\right)\\ =-\log \left(4\right)\end{array}$

Fourth partial sum is:

$\begin{array}{l}=\left[\log \left(1\right)-\log \left(1+1\right)\right]+\left[\log \left(2\right)-\log \left(2+1\right)\right]+\left[\log \left(3\right)-\log \left(3+1\right)\right]+\left[\log \left(4\right)-\log \left(4+1\right)\right]\\ =\left[\log \left(1\right)-\log \left(2\right)\right]+\left[\log \left(2\right)-\log \left(3\right)\right]+\left[\log \left(3\right)-\log \left(4\right)\right]+\left[\log \left(4\right)-\log \left(5\right)\right]\\ =\log \left(1\right)-\log \left(5\right)\\ =0-\log \left(5\right)\\ =-\log \left(5\right)\end{array}$

From the sequence of the partial sums:

$-\log \left(2\right), -\log \left(3\right), -\log \left(4\right), -\log \left(5\right),....$

It is easy to see that the nth partial sum will be $-\log \left(n+1\right)$.

Hence, nth partial sum is $-\log \left(n+1\right)$.