+0

0
212
2

Compute the sum
2/(1•2•3)+2/(2•3•4)+2/(3•4•5)+•••

May 3, 2022

#1
0

sumfor(n, 1, infinity, (2 / (n*(n+1)*(n+2)))==It converges to 1/2

May 3, 2022
#2
+9465
0

The sum, written in sigma notation, is $$\newcommand{\dsum}{\displaystyle\sum} \dsum_{k = 1}^{\infty} \dfrac2{k(k + 1)(k + 2)}$$.

We can perform partial fractions to get a telescoping sum.

Note that $$\newcommand{\dsum}{\displaystyle\sum} \dsum_{k = 1}^\infty (f(k) - f(k + 1)) = f(1) - \lim_{k\to\infty} f(k)$$ for any function f.

Let $$\dfrac2{k(k + 1)(k + 2)} = \dfrac A{k} + \dfrac B{k + 1} + \dfrac C{k + 2}$$ for some constants A, B, C.

Multiplying both sides by k(k + 1)(k + 2) gives $$2 = A(k + 1)(k + 2) + Bk(k + 2) + Ck(k + 1)$$.

Substituting k = 0, k = -1, and k = -2 respectively gives $$\begin{cases}2A = 2\\-B = 2\\2C = 2\end{cases}$$.

Then A = C = 1, B = -2.

We have:

$$\newcommand{\dsum}{\displaystyle\sum} \begin{array}{rcl} \dsum_{k = 1}^{\infty} \dfrac2{k(k + 1)(k + 2)} &=& \dsum_{k = 1}^\infty \left(\dfrac1k + \dfrac1{k + 2} - \dfrac2{k + 1}\right)\\ &=& \dsum_{k = 1}^\infty \left(\dfrac1k - \dfrac1{k + 1}\right) - \dsum_{k = 1} ^\infty\left( \dfrac1{k + 1} - \dfrac1{k + 2}\right) \end{array}$$

Can you continue with my hints here?

May 3, 2022
edited by MaxWong  May 3, 2022