So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 1$ for all $n$. I've been toiling with some algebraic gymnastics for a while now, but I can't seem to get the proof right. Proving it using calculus isn't a problem, but I'm struggling hither.

Why do you think that this problem can be solved using induction? – joriki Jul 24 '12 at 22:08

@joriki the person who gave it to me said it could? So, ethos I guess... – Paquito Jul 24 '12 at 22:10

what about comparing it with $ln2$ – Theorem Jul 24 '12 at 22:17
3 Answers
As often happens with induction proofs, the easiest approach to proving this statement (which doesn't seem inducable at all  after all, how does knowing the sum for $n$ is less than $1$ tell you anything about the sum for $n+1$?) via induction is to transform it into a stronger one: $$\mathrm{For\ all\ } n\geq2, \frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2} \lt 1\frac{1}{n}.$$
Now, the answer becomes a matter of simple algebra:
$$\sum_{i=2}^{n+1} \frac{1}{i^2} = \sum_{i=2}^{n} \frac{1}{i^2} +\frac{1}{(n+1)^2}\lt 1\frac{1}{n}+\frac{1}{(n+1)^2}\lt 1\frac{1}{n}+\frac{1}{n(n+1)} = 1\frac{1}{n+1}.$$
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1Awesome! Many thanks...Might I ask what motivated the choice of subtracting 1/n? – Paquito Jul 24 '12 at 22:21

5@Paquito To a certain extent, intuition  since the difference between consecutive terms of the form $1/n$ is on the order of $1/n^2$, I could see that adding a quadratic term to a $1/n$sized 'gap' between the sum and 1 would give a $1/n+1$sized gap; from there it was just a quick dig to see whether $11/n$ itself would work or whether I would need to do something like $11/(n+1)$. – Steven Stadnicki Jul 24 '12 at 22:26


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@RobertIsrael True enough, and for basically the same reason, but I figured I'd phrase it in discrete terms since the poster was looking for a noncalculus answer. – Steven Stadnicki Jul 25 '12 at 00:16

3@Paquito What Steven did is a classical example of [inventor's paradox](http://en.wikipedia.org/wiki/Inventor%27s_paradox). Similarly proving $\frac{(2n1)!!}{(2n)!!} < \frac{1}{\sqrt{n}}$ does not give in to the method of induction, while a stronger inequality $\frac{(2n1)!!}{(2n)!!} < \frac{1}{\sqrt{n+1}}$ readily does. The explanation for why this works, is that the assumptions of the induction step are stronger in the latter case, allowing to draw stronger conclusion. – Sasha Jul 25 '12 at 01:22

See also: http://math.stackexchange.com/questions/1150388/summationinductionalprooffrac112frac122frac132ldots – Martin Sleziak Apr 30 '15 at 12:47
Another proof is by comparison: note that
$$ {1 \over k^2} < {1 \over (k1)k} $$
for all integers $k \ge 2$. Therefore
$$ {1 \over 2^2} + {1 \over 3^2} + \cdots + {1 \over n^2} < {1 \over 1 \times 2} + {1 \over 2 \times 3} + \cdots + {1 \over (n1) \times n} $$
and now you need to find the sum on the righthand side. But you can actually write
$$ {1 \over (k1)k} = {1 \over k1}  {1 \over k} $$
(this is just the usual partial fraction decomposition) and therefore
$$ {1 \over 1 \times 2} + {1 \over 2 \times 3} + \cdots + {1 \over (n1) \times n} = \left( {1 \over 1}  {1 \over 2} \right) + \left( {1 \over 2}  {1 \over 3} \right) + \cdots + \left( {1 \over n1}  {1 \over n} \right) $$
and the righthand side what's called a ``telescoping sum''  that is, the pairs of terms $1/2$ and $+1/2$, $1/3$ and $+1/3$, and so on cancel. So the righthand side is $1  1/n$, which is less than 1.
This came to mind pretty much immediately for me, because I happened to know that $\sum_{k \ge 2}^\infty 1/(k(k1)) = 1$, but if you didn't know that ahead of time it would be a bit tricky to discover.
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Yet another approach :
Let us first analyze the sum till infinity. Let $$ S= \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$ $$\Rightarrow S=(\frac{1}{2^2}+\frac{1}{4^2}+ \cdots\infty) +(\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty ) $$$$\Rightarrow S= \frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\infty)+ S' $$ Where $$S'=\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty $$ $$\Rightarrow S=\frac{1}{4}(1+S)+S'$$ $$\Rightarrow 3S=4S'+1........ Eqn(1)$$ Now examine the following inequality $$ (\frac{1}{2^2}\frac{1}{3^2})+(\frac{1}{4^2}\frac{1}{5^2})+ \cdots \infty > 0$$ $$\Rightarrow \frac{1}{2^2}+\frac{1}{4^2}+\cdots > \frac{1}{3^2}+\frac{1}{5^2}+ \cdots$$ $$\Rightarrow \frac{1}{4}(1+\frac{1}{2^2}+\cdots) >\frac{1}{3^2}+\frac{1}{5^2}+ \cdots $$ $$\Rightarrow \frac{1}{4}(1+S)> S'$$ $$\Rightarrow (1+S)> 4S'......Eqn(2)$$ From Equation 1 and 2 we get $$ 1+S> 3S1$$ $$\Rightarrow 1> S$$ Which shows $$ 1> \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$
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