Question

How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given `1/(1*2)+1/(2*3)+...+1/(n(n+1))+...`?

Answer 1

`sum_(n=1)^oo 1/(n(n+1)) = 1`

Explanation:

Given the series:

`sum_(n=1)^oo 1/(n(n+1))`

we can determine that is convergent using the direct comparison test since:

`1/(n(n+1)) < 1/n^2`

and the series:

`sum_(n=1)^oo 1/n^2`

is convergent based on the `p`-series test.

Decompose now the general term using partial fractions:

`1/(n(n+1)) = 1/n-1/(n+1)`

and note that in the partial sum:

`s_n = (-1/(n+1) +1/n) + (-1/n +1/(n-1)) + ... + (-1/3+1/2) + (-1/2 +1)`

all terms cancel each other except the first and the last, so:

`s_n = 1-1/(n+1)`

and then:

`lim_(n->oo) s_n = 1`

is the sum of the series.

Answer 2

The answer is `=1`

Explanation:

`sum_(k=1)^n1/(k(k+1))=sum_(k=1)^n1/k-sum_(k=1)^n1/(k+1)`

`=sum_(k=1)^n1/k-sum_(k=2)^(n+1)1/(k)`

`=1+cancel(sum_(k=2)^n1/k)-cancel(sum_(k=2)^n1/k)-1/(n+1)`

`=1-1/(n+1)`

`=n/(n+1)`

`lim_(n->+oo)n/(n+1)=lim_(n->+oo)1/(1+1/n)=1`

As

`lim_(n->+oo)1/n=0`