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#

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