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Mathematics 43 Online
OpenStudy (anonymous):

c) Prove, that the series \(\sum_{n=1}^{\infty}\frac{1}{(n+1)(n+2) }\) converges and calculate the value of the series Hint: Telescoping sum.

OpenStudy (anonymous):

\[ \frac{1}{(n+1) (n+2)}=\frac{1}{n+1}-\frac{ 1}{n+2}\\ s_n=\\ \frac{1}{2}-\frac 1 3+\\ \frac{1}{3}-\frac 1 4+\\ \frac{1}{4}-\frac 1 4+\\ \cdots\\ \cdots\\ \frac{1}{n+1}-\frac{ 1}{n+2}\\ s_n =\frac 1 2 -\frac{ 1}{n+2}\\ \lim_{n\to \infty} s_n=\frac 1 2 \]

OpenStudy (anonymous):

thank you very much dear mr Elias, there is some parts that i dont understand but i will ask later, i must first ensure points for tomorow :)

OpenStudy (anonymous):

\[ \sum _{n=1}^{\infty } \frac{1}{(n+1) (n+2)}=\frac{1}{2} \]

OpenStudy (anonymous):

and its not convergence right ? its divergence

OpenStudy (anonymous):

No. It is convergent and its sum is 1/2

OpenStudy (anonymous):

hm ok.. thanks

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