I have an infinite serie sum from 1 to infinite of 1/( n*(n+1) ). How can i found out if this converge ?

Question

anonymous · Answer

Of course it does BECAUSE ...

anonymous · Answer

There is COMPARISON TEST - YOUR series each summand  is smaller than 1/$$n^{2}$$ but it is well known that $$\sum_{1}^{\infty} \frac{ 1 }{ n^{2} }$$ IS CONVERGENT

anonymous · Answer

As U realize n* (n+1) $$n(n+1)$$<  $$n^{2}$$

anonymous · Answer

Sorry I meant the sign Bigger n(n+1) > n^2

anonymous · Answer

So their Inverses are in the opposite Relation !

anonymous · Answer

Do You understand this argument ?

anonymous · Answer

n(n+1) is bigger than n^2 so 0 < n^2 < n(n+1). Now with their inverses it reverses it. 0 < 1/[n(n+1)] < 1/[n^2] ? I know that 1/n^2 is convergent so by comparison test I can afirm that 1/ [n(n+1)] is also convergent ? I understood rigth ?

anonymous · Answer

Yes U DID UNDERSTAND RIGHT. But...

anonymous · Answer

Writing the answer in such "simplistic" form will show that you do not fully grasp the question. You Should A. READ THE f.-g  EXACT THEOREM OF COMAPRISON OF SERIES.
B.Understand the importance of "FOR ALL n" in its formulation.
C. Stress "for all n" in your written answer

anonymous · Answer

you can also find the sum if you like

anonymous · Answer

satellite - this is actually contrary to the best-interest of the student - since he should LEARN the COMPARISON TEST and not some special case which may be USELESS in other examples !!!

anonymous · Answer

since $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ your series telescopes

anonymous · Answer

Satll. No offense intended - just pedagogy