Find the general term of the sequence, starting with n=1 Find the general term of the sequence, starting with n=1, determine whether the sequence converges, and if so ﬁnd its limit.
$$\frac{3}{2^2 - 1^2}, \frac{4}{3^2 - 2^2} , \frac{5}{4^2 - 3^2}, \cdots$$
Can you help me with this,I know how to solve the problem with n= 0 but is that different with n= 1? 
 A: Observe the series first.
$$  \frac{1\color{green}{+2}}{(1\color{green}{+1})^2-(1)^2}, \frac{2\color{green}{+2}}{(2\color{green}{+1})^2-(2)^2}, \frac{3\color{green}{+2}}{(3\color{green}{+1})^2-(3)^2}, .... 
 \text{upto} \frac{n\color{green}{+2}}{(n\color{green}{+1})^2-(n)^2} $$
So now,
General Term or $ n^{th} $ term can be written as:
$$ \frac{(n+2)}{(n+1)^2-n^2} $$
$$ \frac{(n+2)}{(\require{cancel} \cancel{n^2}+2n+1-
 \cancel{n^2}) }$$
$$ = \frac{(n+2)}{(2n+1)}$$
A: Note that the general formula for the $n$th term is: $$T_n = \frac{n+2}{(n+1)^2-n^2}$$ which can be simplified using difference of squares ($a^2-b^2=(a-b)(a+b)$) as: $$T_n = \frac{n+2}{2n+1}$$
A: General term:
$$\frac{n+2}{(n+1)^2-n^2}$$
That is,
$$\frac{n+2}{2n+1}$$
Can you find the limit now?
A: The expressions above proves divergence by the nth term test, i.e. the limit of the sequence does not go to zero.
Here's a small survey of other techniques. 
Is interesting to note that this sequence leads to an inconclusive Ratio Test, i.e.  |T_n+1/T_n|=1. 
Let (n+2)/(2*n+1)=T_n
Then 2*T_n =  (2*n+4)/(2*n+1) =   1 + 3/(2*n+1).
So, T_n = (1/2) + 3/[2*(2*n+1)].
This expression has a somewhat easier limit to evaluate, however to get there required extra steps. From here though, its easier to approximate the sum. 
The constant term, (1/2), guarantees each additional term in the sum adds a half, so part of the Partial Sum is (N/2). 
The second term is  (3/2)/(2*n+1)= (3/4)/(n+1/2)> (3/4)/(n+1), 
so sum(T_n) > (N/2) + (3/4)*sum(1/(k+1), k, 1,N), with the right hand side being a lower bound for the sum.
The second sum is approximately the natural log of N.
So sum(T_n,1,n) > (N/2) + (3/4)*ln(N/e). 
