1. ## telescopeing series?

i dont understand how i tell if a telescopeing series is convergent or divergent...

problem:
1/n(n+1)

i used partial fractions and got it down to this....

S_n = 1 - 1/(n+1) = total sum = 1

what do i look at/do to know if this is convergent or divergent

2. Originally Posted by lochnessmonster
i dont understand how i tell if a telescopeing series is convergent or divergent...

problem:
1/n(n+1)

i used partial fractions and got it down to this....

S_n = 1 - 1/(n+1) = total sum = 1

what do i look at/do to know if this is convergent or divergent

The series $\displaystyle \sum_{k=1}^{\infty}a_k$ is converges if there exist a finite limit $\displaystyle S$ of partial sum sequence $\displaystyle \{S_n\}$ and $\displaystyle S$ is the sum of the series $\displaystyle \sum_{k=1}^{\infty}a_k$.
If the limit $\displaystyle \{S_n\}$ isn't exist, we say that the series is divergent.

Now, to telescopic series.

$\displaystyle \sum_{n=1}^{\infty}(a_{n+1}-a_n)$ is telescopic series.

Partial sum $\displaystyle S_n$ of $\displaystyle \sum_{n=1}^{\infty}(a_{n+1}-a_n)$ is:

$\displaystyle S_n=(a_2-a_1)+(a_3-a_2)+...+(a_{n+1}-a_n)=a_{n+1}-a_1$

By the definition above $\displaystyle \sum_{n=1}^{\infty}(a_{n+1}-a_n)$ converges if $\displaystyle lim_{n\to\infty}S_n$ exist.

Hence, $\displaystyle lim_{n\to\infty}S_n=lim_{n\to\infty}a_{n+1}-a_1$, so $\displaystyle \sum_{n=1}^{\infty}(a_{n+1}-a_n)$ converges iff $\displaystyle \{a_n\}$ converges.

$\displaystyle \sum_{n=1}^{\infty}(\frac{1}{n(n+1)}=\sum_{n=1}^{\ infty}(\frac{1}{n}-(\frac{1}{n+1}$
$\displaystyle \{\frac{1}{n}\}$ is converges sequence, so the series is also converges.
$\displaystyle S_n=1-\frac{1}{n+1}$ and $\displaystyle lim_{n\to\infty}S_n=1$, hence $\displaystyle \sum_{n=1}^{\infty}(\frac{1}{n(n+1)}=1$