How do I write #(5/(1*2))+(5/(2*3))+(5/(3*4))+...+(5/n(n+1))+...#in summation notation, and how can I tell if the series converges?

2 Answers
Harish Chandra Rajpoot
Jul 19, 2018

#5\sum_{k=1}^n\frac{1}{k(k+1)}#

Explanation:

The given series:

#(5/{1\cdot 2})+(5/{2\cdot 3})+(5/{3\cdot 4})+\ldots+(5/{n(n+1)})#

#=\sum_{k=1}^n\frac{5}{k(k+1)}#

#=5\sum_{k=1}^n\frac{1}{k(k+1)}#

#=5\sum_{k=1}^n(1/k-\frac{1}{k+1})#

#=5((1-1/2)+(1/2-1/3)+(1/3-1/4)+\ldots+(1/n-1/{n+1}))#

#=5(1-1/2+1/2-1/3+1/3-1/4+\ldots+1/n-1/{n+1})#

#=5(1-1/{n+1})#

#\therefore \lim_{n\to \infty}\sum_{k=1}^n\frac{5}{k(k+1)}#

#=\lim_{n\to \infty}5(1-1/{n+1})#

#=5(1-0)#

#=5#

Hence, the given series is converging

Narad T.
Jul 19, 2018

Please see some comlementary details below.

Explanation:

I added a few more details

The partial fraction decomposition is

#1/(k(k+1))=A/k+B/(k+1)#

#=(A(k+1)+Bk)/(k(k+1))#

Compare the numerators

#1=A(k+1)+Bk#

Let #k=0#, #=>#, #1=A#

Let #k=-1#, #=>#, #1=-B#

Therefore,

#1/(k(k+1))=1/k-1/(k+1)#

Therefore,

#sum_(k=1)^n1/(k(k+1))=sum_(k=1)^n1/k-sum_(k=1)^n1/(k+1)#

#=1+sum_(k=2)^n1/k-(sum_(k=1)^(n-1)1/(k+1))+1/(n+1)#

#=1+sum_(k=2)^n1/k-sum_(k=2)^(n)1/(k)-1/(n+1)#

#=1-1/(n+1)#

#=n/(n+1)#

#=1/(1+1/n)#

And

#lim_(n->oo)sum_(k=1)^n1/(k(k+1))=lim_(n->oo)1/(1+1/n)#

#=1#

The series converges and the limit is #=5#

Related questions
See all questions in Strategies to Test an Infinite Series for Convergence
Impact of this question
4554 views around the world
You can reuse this answer
Creative Commons License