# telescopeing series?

• Mar 23rd 2011, 01:10 PM
lochnessmonster
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
• Mar 23rd 2011, 01:44 PM
Also sprach Zarathustra
Quote:

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

Now, to telescopic series.

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

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

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

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

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

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