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**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.