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?
Best Answer
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:
$$ \frac{(n+2)}{(n+1)^2-n^2} $$
$$ \frac{(n+2)}{(\require{cancel} \cancel{n^2}+2n+1- \cancel{n^2}) }$$
$$ = \frac{(n+2)}{(2n+1)}$$