Solving $Т(n)=\frac{1}{\frac{1}{T(n-1)}+n^2}$ and $T(1) = 1$

Question

$$T(n) = \frac1{\frac1{T(n-1)}+n^2}$$

I did find $T(2) = \frac{1}{1+n^2}$, but I don't know how to proceed.

How do I go on from here? How can I find the solution? Thanks. :)

Answer 1

$$\frac{1}{t_{n}}=\frac{1}{t_{n-1}}+n^2$$ and the rest is smooth I think:

$$\frac{1}{t_2}=\frac{1}{t_1}+2^2$$ $$\frac{1}{t_3}=\frac{1}{t_2}+3^2...$$ $$\frac{1}{t_n}=\frac{1}{t_{n-1}}+n^2.$$ A summing of these equalities gives:
$$\frac{1}{t_n}-\frac{1}{t_1}=2^2+3^2+...+n^2$$ or $$\frac{1}{t_n}=1^2+2^2+3^2+...+n^2$$ or $$\frac{1}{t_n}=\frac{n(n+1)(2n+1)}{6}.$$ The answer is: $$t_n=\frac{6}{n(n+1)(2n+1)}$$