For each series, find an explicit formula for the sequence of partial sums and determine if the series converges. sum_{n=1}^inftyfrac{1}{n(n+1)}

Given: The series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}$
To determine: The explicit formula for the sequence of partial sum and convergence of the series.
Explanation:
The general term (k-th term) of the sequence of partial sum is of a series is the sum of first k terms.
Here the given series is $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}$
So general term of sequence of partial sum of this series is $\sum _{n=1}^{n=k}\frac{1}{n(n+1)}$
This general can be re-written as
$t_k=\sum _{n=1}^{n=k}\frac{1}{n(n+1)}$
$\Rightarrow t_k=\sum _{n=1}^{n=k}(\frac{1}{n}-\frac{1}{n+1})$
$\Rightarrow t_k=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{k}-\frac{1}{k+1})$
$\Rightarrow t_k=(1-\frac{1}{k+1})$
Hence general term of sequence of partial sum of this series is $t_k=(1-\frac{1}{k+1})$
So sequence of partial sum is ${\{1-\frac{1}{n+1}\}}_{n=1}^{\mathrm{\infty }}$
Now it is known that $\sum _{n=1}^{\mathrm{\infty }}{a}_n=\underset{k\to \mathrm{\infty }}{lim}\sum _{n=1}^k{a}_n$
Therefore here
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}=\underset{k\to \mathrm{\infty }}{lim}\sum _{n=1}^k\frac{1}{n(n+1)}$
$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}=\underset{k\to \mathrm{\infty }}{lim}(1-\frac{1}{k+1})$
$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}=1$
Hence the series is convergent as converges to 1.
Answer:
The sequence of partial sum is ${\{1-\frac{1}{n+1}\}}_{n=1}^{\mathrm{\infty }}$
And the series is convergent as converges to 1.