<s>[INST] induction clarification about the step $n+1$ Suppose i need to prove that $\frac{1}{2^2}+\frac{1}{3^2}...+\frac{1}{n^2}<1-\frac{1}{n}$
So in the step of $n+1$, the right side becomes $<1-\frac{1}{n+1}$ or is it: $<1-\frac{1}{n}-\frac{1}{n+1}$?  i guess it's the first one but why? i mean in the left side it's in addition to the first term $(\frac{1}{n^2})$ which is $\frac{1}{(n+1)^2}$. 
*This question is mostly because i saw answers for this where they add $\frac{1}{(k+1)^2}$ to both sides, which is complex than just to write : $1-\frac{1}{n}=\frac{n-1}{n}$ and in the step of $n+1$ it just becomes: $\frac{n}{n+1}$ from there it's simple so i didn't understand why to add that term unless i'm missing something...
Thank you!
Thanks you.
 [/INST] Your induction should be as follows: assume that this property holds for all $k≤n$. Then your inductive step for $n+1$ is: $$\frac{1}{3^2} +\ldots +\frac{1}{n^2}+\frac{1}{(n+1)^2}<\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}$$ This holds as you can verify for yourself that $\frac{1}{2^2}>\frac{1}{(n+1)^2}$ for any $n≥2$. Since $\frac{1}{n}>\frac{1}{n+1}$, and by our assumption, we now have that the equation is: $$\frac{1}{3^2} +\ldots +\frac{1}{(n+1)^2}< \frac{1}{2^2}+\ldots +\frac{1}{n^2}<1-\frac{1}{n}<1-\frac{1}{n+1}$$ And our inductive step is shown.
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