What is #1/3+1/8+1/15+1/24+1/35+1/48+...+1/(n^2+2n)+...# ?

1 Answer
George C.
Jan 14, 2018

#3/4#

Explanation:

The general formula of a term of this series can be written:

#a_n = 1/(n^2+2n) = 1/2(1/n-1/(n+2))#

So:

#sum_(n=1)^N a_n = 1/2 sum_(n=1)^N (1/n-1/(n+2))#

#color(white)(sum_(n=1)^N a_n) = 1/2 (sum_(n=1)^N 1/n- sum_(n=1)^N 1/(n+2))#

#color(white)(sum_(n=1)^N a_n) = 1/2 (sum_(n=1)^N 1/n- sum_(n=3)^(N+2) 1/n)#

#color(white)(sum_(n=1)^N a_n) = 1/2 (sum_(n=1)^2 1/n+color(red)(cancel(color(black)(sum_(n=3)^N 1/n))) - color(red)(cancel(color(black)(sum_(n=3)^N 1/n)))- sum_(n=N+1)^(N+2) 1/n)#

#color(white)(sum_(n=1)^N a_n) = 1/2 (1/1+1/2-1/(N+1)-1/(N+2))#

Hence:

#sum_(n=1)^oo a_n = lim_(N->oo) 1/2 (1/1+1/2-1/(N+1)-1/(N+2)) = 3/4#

Related questions
Impact of this question
6111 views around the world
You can reuse this answer
Creative Commons License