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I am having trouble with the following proof:

For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$

My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}{k}$ in the inductive step and reduce it down to $2-\frac{1}{k+1}$ but cannot do so. I am beginning to think that my entire approach is wrong.

Martin Sleziak
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mrQWERTY
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  • $$n\lt n+1\iff n(n+1)\lt (n+1)^2\iff \dfrac{1}{(n+1)^2}\lt\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$$ – Bumblebee Mar 25 '15 at 05:35
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    See http://math.stackexchange.com/questions/1251544/proof-of-sum-k-1n-frac1k2-2-frac1n and http://math.stackexchange.com/questions/1220203/proving-1-frac14-frac19-cdots-frac1n2-leq-2-frac1n-for – Martin Sleziak Oct 26 '15 at 14:50

5 Answers5

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Hint: $\frac{1}{n^2} < \frac{1}{(n-1)} - \frac1n$, and telescope.

MT_
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DeepSea
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Inductive Hypothesis: suppose $\sum_{k=1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$.

Inductive Step: then suppose $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq 2 - \frac{1}{n} + \frac{1}{(1+n)^2}. $

So it suffices to show that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$.

But this, by simple algebra, is equivalent to $ n \leq n+1 $, which is obviously true $\forall n \in \mathbb{N}$.

EDIT Proof that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$ is equivalent to $n \leq n+1$.

Add $\frac{1}{n}$ to both sides to se that what we need to show is $\frac{1}{(1+n)^2} \leq \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}.$ Multiply both sides by $(n+1)$. Then take reciprocals and remember to swap the inequality sign.

Frank
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In the induction step observe that $1$+$1/4$ +...+$1/k^{2}$ + $1/(k+1)^{2}$$\leq$ $2$- $1/k$ + $ 1/(k+1)^{2}$

Now $1/(k+1)^{2}$ $\leq$ $1/k(k+1)$=$1/k$-$1/(k+1)$. Use this inequality in previous to obtain the answer.

wanderer
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Others have commented on using induction. Here's another approach just 'cause:

We know the infinite sum is equal to $\frac{\pi^2}{6}$ which is approximately $1.644$, so we know it is true for $n > 2$ (infinite sum > partial sum with all positive terms)

So we can just hand check $n = 1, 2$: Indeed, $1 \leq 1$ and $\frac{5}{4} \leq \frac{3}{2}$

MT_
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  • talk about sledgehammers and molehills! – Frank Mar 19 '14 at 21:28
  • it was never said that you couldn't use that and the two questions aren't really that related so figured why not! – MT_ Mar 19 '14 at 21:30
  • you could probably use Fermat's theorem to prove lots of elementary things ;) – Frank Mar 19 '14 at 21:31
  • haha, yeah. There was a recent homework post to prove that $(a+1)^3 \neq a^3 + (a-1)^3$ for any positive integer $a > 2$, and someone jokingly commented with a wikipedia article to FLT. – MT_ Mar 19 '14 at 21:33
  • hahahahah excellent :) – Frank Mar 19 '14 at 21:34
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Just thought I should comment on the inequality $$-\frac{1}{n} + \frac{1}{(n+1)^2} \leq -\frac{1}{n+1}$$ that needs to be shown in the induction argument. If we let $f(x) = 1/x$ then this is equivalent to showing $f(n+1)-f(n) \leq f'(n+1)$. But $$f(n+1)-f(n) = \int_n^{n+1} f'(x) dx \leq f'(n+1)$$ since $f'(x) = -1/x^2$ is increasing for positive $x$.

abnry
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  • It doesn't need to be shown in any sophisticated way at all. As I show, it is immediately equivalent to the inequality $n \leq n+1$. – Frank Mar 19 '14 at 22:59
  • Nothing needs to be shown in any way at all! I'm just giving another solution. (Besides, it's only calculus.) From this we could generalize the above problem to proving that $\sum_{k=1}^n f'(k) \leq C + f(n)$ for an appropriate constant $C$, or something like that, where $f'(x)$ is increasing. – abnry Mar 19 '14 at 23:50