Is-the-following-series-absolutely-convergent-S-1-n-1-1-n-n-1-Is-the-following-series-absolutely-convergent-S-2-n-1-1-n-1-n-1-

Question Number 2045 by prakash jain last updated on 31/Oct/15

I s t h e f o l l o w i n g s e r i e s a b s o l u t e l y c o n v e r g e n t ?

S_{1} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + 1)}

I s t h e f o l l o w i n g s e r i e s a b s o l u t e l y c o n v e r g e n t ?

S_{2} = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1})

Commented by 123456 last updated on 31/Oct/15

\frac{1}{n (n + 1)} = \frac{1}{n} - \frac{1}{n + 1}

its telescoping serie and it converge since

\underset{i = 1}{\sum^{n}} (\frac{1}{i} - \frac{1}{i + 1}) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots + \frac{1}{n} - \frac{1}{n + 1}

= 1 - \frac{1}{n + 1} \to 1 (n \to + \infty)

Commented by prakash jain last updated on 31/Oct/15

So S_{1} = S_{2} and convergent . Then it has to be absolutely

convergent, since each term is + ve .