Is the series 1/(n(n+1)) convergent?
n=2 to n=infinity
To determine if the series 1/(n(n+1)) is convergent or divergent, we can use the concept of telescoping series. A telescoping series is a series in which many of the terms cancel out, leaving only a finite number of terms to be summed.
Let's start by writing out the terms of the series:
1/[(2)(2+1)] + 1/[(3)(3+1)] + 1/[(4)(4+1)] + ...
Now, let's simplify the terms:
1/[2(3)] + 1/[3(4)] + 1/[4(5)] + ...
Notice that the denominator of each term can be rewritten as (n)(n+1). Now, let's rewrite the terms in a telescoping form:
(1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + ...
We can observe that every term cancels out except for the first and the last term. So, the series reduces to:
1/2 - 1/n+1
Now, let's find the limit of this expression as n approaches infinity to determine if the series converges:
lim (nāā) [1/2 - 1/(n+1)]
As n approaches infinity, the second term, 1/(n+1), will approach zero. Therefore, the limit becomes:
lim (nāā) (1/2 - 0)
Which is 1/2.
Since the limit is a finite number, the series 1/(n(n+1)) converges. Specifically, it converges to 1/2.