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Reading this proof, I don't understand a part of the inductive step (copied below). I do not understand how we are going from

$$ \leq 2-\frac{1}{k}+\frac{1}{(k+1)^2}\quad\text{(by $S(k)$)} $$ to $$ = 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right) $$

Inductive step: Fix some $k\geq 2$ and suppose that $S(k)$ is true. It remains to show that $$ S(k+1) : 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}\leq 2-\frac{1}{k+1} $$ holds. Starting with the left-hand side of $S(k+1)$, \begin{align} 1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2} &\leq 2-\frac{1}{k}+\frac{1}{(k+1)^2}\quad\text{(by $S(k)$)}\\[1em] &= 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)\\[1em] &= 2-\frac{1}{k+1}\left(\frac{k^2+k+1}{k(k+1)}\right)\tag{simplify}\\[1em] &< 2-\frac{1}{k+1}.\tag{$\dagger$} \end{align}

Robert W
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    If you multiply out $-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)$ you will see that you get $-\frac{1}{k}+\frac{1}{(k+1)^2}$. – rogerl Apr 19 '21 at 00:11
  • @rogerl thank you. How do you think they figured out how to expand it that way? Do I just need to read up more on manipulating fractions? – Robert W Apr 19 '21 at 00:16
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    This proof obscures what is going on. It is hiding that, for $n>1$ $$\frac1{n^2}<\frac1{n(n-1)}=\frac1{n-1}-\frac1n$$ and then using the right side to get a “telescoping sum.” This is probably where the proof came from, originally. – Thomas Andrews Apr 19 '21 at 00:33
  • @ThomasAndrews Okay, that makes a ton of sense. Thank you. – Robert W Apr 19 '21 at 01:12

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