This is a seemingly simple induction question that has me confused about perhaps my understanding of how to apply induction

the question;

$$\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\ \le\ 2-\frac{1}{n},\ \forall\ n \ge1.$$

this true for $n=1$, so assume the expression is true for $n\le k$. which produces the expression,

$$\frac{1}{1^2} + \cdots + \frac{1}{k^2} \le\ 2-\frac{1}{k}.$$ now to show the expression is true for $k+1$,

$$\frac{1}{1^2}+\cdots+ \frac{1}{k^2} + \frac{1}{(k+1)^2} \le\ 2-\frac{1}{k}+\frac{1}{(k+1)^2}.$$

this the part I am troubled by, because after some mathemagical algebraic massaging, I should be able to equate,


which would prove the expression is true for $k+1$ and I'd be done. right? but these two are not equivalent for even $k=1$, because setting $k=1$ you wind up with $\frac{5}{4}=\frac{3}{2}$, so somewhere i am slipping up and I'm not sure how else to show this if someone has some insight into this induction that I'm not getting. thanks.


marked as duplicate by Martin Sleziak, Najib Idrissi, drhab, Davide Giraudo, kingW3 Feb 2 '15 at 10:20

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    $\begingroup$ If you can't get $2-\frac{1}{k}+\frac{1}{(k+1)^2}=2-\frac{1}{(k+1)}$, the next natural thing is to try to get $2-\frac{1}{k}+\frac{1}{(k+1)^2}\leq2-\frac{1}{(k+1)}.$ $\endgroup$ – Git Gud Jul 16 '14 at 21:22
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    $\begingroup$ You need $$2 - \frac{1}{k} + \frac{1}{(k+1)^2} \leqslant 2 - \frac{1}{k+1},$$ not equality. $\endgroup$ – Daniel Fischer Jul 16 '14 at 21:22
  • $\begingroup$ And the last equality is equivalent to $\frac1{(k+1)^2} \le \frac1k-\frac1{k+1}$. After using common denominator on the RHS, this inequality should be clear. $\endgroup$ – Martin Sleziak Jul 27 '14 at 6:19

What you really need is $2 − \frac{1}{k} + \frac{1}{(k+1)^2} \leq 2 − \frac{1}{(k+1)}$,

  • $\begingroup$ ok so is it enough to say that the inequality $$2-\frac{1}{k}+\frac{1}{(k+1)^2} \le\ 2-\frac{1}{(k+1)}$$ is equivalent to the inequality $$\frac{1}{(k+1)}+\frac{1}{(k+1)^2} \le\ \frac{1}{k}$$, which is obvious enough to see? $\endgroup$ – user74091 Jul 16 '14 at 21:36
  • $\begingroup$ what i mean to say is that the truth of the second inequality is obvious enough to see, albeit i could do an induction proof of it as well for sake of completion. $\endgroup$ – user74091 Jul 16 '14 at 21:45
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    $\begingroup$ Well, multiplying by $k(k+1)^2$, the inequality is seen to be equivalent to $k^2 + 2k \leq k^2 +2k +1$, which is now certainly obvius. $\endgroup$ – Marco Flores Jul 16 '14 at 21:46
  • $\begingroup$ oh, yes. well that is certainly better than either waving my hands and saying that the truth of the second inequality is obvious, or proving the second inequality by induction again. lol. thanks again. $\endgroup$ – user74091 Jul 16 '14 at 21:51

Despite the fact that such solution is given in other posts (for example in this question about similar infinite series: Proof that $\sum_{1}^{\infty} \frac{1}{n^2} <2$ it is used as an auxiliary result in some answers) it might be useful to mention the solution using telescoping sum in a question asking about finite sum.

Let us look at the sum starting with $k=2$: $$\sum\limits_{k=2}^n \frac1{k^2} \le \sum\limits_{k=2}^n \frac1{k(k-1)}.$$ After rewriting this as $\frac1{k(k-1)}=\frac1{k-1}-\frac1k$we see that on the RHS we get a telescoping sum, where many terms will cancel out: $$\sum_{k=2}^n \left(\frac1{k-1}-\frac1k\right)=(1-\frac12)+\left(\frac12-\frac13\right)+\dots+\left(\frac1{n-1}-\frac1n\right)=1-\frac1n$$ (We started with $k=2$ so that we do not get zero in the denominator in the expressions $\frac1{k-1}$ and $\frac1{k(k-1)}$.)


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