
Originally Posted by
dwally89
Is this correct?
Proof by induction
Prove for $\displaystyle n=2$
$\displaystyle a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2$
$\displaystyle 1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2$
Case holds for $\displaystyle n=2$
Assume case holds for $\displaystyle n=q$, i.e.
$\displaystyle a_q\le1+\sum_{k=1}^{q-1} \frac{1}{k^2}$
Prove for $\displaystyle n=q+1$
$\displaystyle a_{q+1}=a_q+\frac{1}{n^2a_q}$
$\displaystyle a_{q+1}\le 1+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2a_q}$
$\displaystyle a_{q+1}\le a_{q+1}+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}$
$\displaystyle 0\le \sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}$
which is clearly true.