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Reference: https://i.imgur.com/xlD2aI9.png

I can see why $\frac{-1}{k}+\frac{1}{(k+1)^2} \leq \frac{-1}{k+1}$ holds: multiplying both sides of the inequality by $(k+1)$ allows me to simplify to the form $\frac{1}{k+1} \leq \frac{1}{k}$, but it's unclear to me why $\frac{-1}{k}+\frac{1}{(k+1)^2} \leq \frac{-1}{k+1}$ is sufficient to show that $\sum_{i=1}^ki + \frac{1}{(k+1)^2} \leq \ 2-\frac{1}{k}+\frac{1}{(k+1^2)}$.

Did
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greedIsGoodAha
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  • The title was absurd and I modified it. Remaining are some serious typos in the body. Also, a reference of the book the page this is asking about is taken from should be given. – Did Feb 11 '17 at 12:29

1 Answers1

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The thing is, we do not want to show that $$\sum_{i=1}^k \frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k}+\frac{1}{(k+1)^2}.$$

Instead, the thing we must show in the inductive case is $$ \sum_{i=1}^k \frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k+1}. \tag1 $$

Prior to the point where you had your doubts, the proof had already shown that $$ \sum_{i=1}^k \frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2 + \left( -\frac{1}{k}+\frac{1}{(k+1)^2}\right). \tag2 $$ Can you see how Inequality $(2)$, plus the fact that $-\dfrac{1}{k}+\dfrac{1}{(k+1)^2} \leq -\dfrac{1}{k+1}$, is enough to prove Inequality $(1)$?

Martin Sleziak
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David K
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  • Is it because $2+\frac{-1}{k+1} \leq 2+(\frac{-1}{k}+\frac{1}{(k+1)^2})$? That makes sense, but only if we know that the summation of $\frac{1}{i^2}$ from $i=1$ to $k$ is less than $2$ to begin with. Why do we know that? – greedIsGoodAha Nov 12 '16 at 00:04
  • No, you have the inequality backward in that comment. Oh, and by the way, in the inductive step we do know that the sum of $1/i^2$ from $1$ to $k$ is less than $2$, in fact less than $2-1/k$. That was the inductive assumption. The inductive assumption is always justified during the inductive step. That is how induction works. – David K Nov 12 '16 at 00:11