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I am trying to solve this inequality by induction. I just started to learn induction this week and all the inequalities we had been solved were like an equation less than another equation (e.g. $n! \geq 2^n$) So I am confused how to prove an inequality that is less than a particular value like the following problem? Thanks in advance.

$$\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\ldots+\frac1{n^2}<2$$

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marked as duplicate by Martin Sleziak, SchrodingersCat, Martin R, Nehorai, Alex M. Jan 17 at 13:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Can you show $\text{LHS} < 2 - f(n)$ where $f(n) \rightarrow 0$? This should get it in the shape you are familiar with. – Eric Towers Jan 17 at 3:58
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You are going to have a hard time proving it by induction, because know $s_n<2$ is not going to help you know $s_n+a_{n+1}<2$. Rather, you need a variable bound, like $s_n<2-f(n)$ for some function $f$ which is known to be positive so that $f(n)-a_{n+1}<f(n+1)$... – Thomas Andrews Jan 17 at 4:02
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See also this question and also other related posts. – Martin Sleziak Jan 17 at 7:19
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Other closely related posts can be found here and here. (And again, you can have a look at the linked questions.) – Martin Sleziak Jan 17 at 12:00
    
This post is also related to your question: Illustrative examples of a phenomenon in the logic of mathematical induction – Martin Sleziak Jan 17 at 13:16
up vote 8 down vote accepted

A useful fact about proofs by induction is that sometimes it’s easier to prove a stronger result. That happens to be the case here. Let $$s_n=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\ldots+\frac1{n^2}\;.$$

Then

$$\begin{align*} 2-s_1&=1\ge 1\;,\\ 2-s_2&=\frac34\ge\frac12\;,\\ 2-s_3&=\frac34-\frac19=\frac{23}{36}\ge\frac13\;,\text{ and}\\ 2-s_4&=\frac{23}{36}-\frac1{16}=\frac{83}{144}\ge\frac14\;. \end{align*}$$

Those numbers $1,\frac34,\frac{23}{36}$, and $\frac{83}{144}$ are the ‘amounts of space’ left between $s_1,s_2,s_3$, and $s_4$ on the one hand, and the bound of $2$ on the other. They don’t seem to be shrinking very fast: $\frac{83}{144}$ is still $0.5763\overline{8}$. It looks as if they might be shrinking slower than the simple sequence $1,\frac12,\frac13,\frac14,\ldots\;$; certainly they are so far, and we have

$$\begin{align*} s_1&=1\le 2-1\;,\\ s_2&\le 2-\frac12\;,\\ s_3&\le 2-\frac13\;,\text{ and}\\ s_4&\le 2-\frac14\;. \end{align*}$$

This suggests that just maybe $s_n\le 2-\dfrac1n$ for all $n\ge 1$. If true, that would certainly imply that $s_n<2$.

HINT: Try to prove by induction that $s_n\le 2-\dfrac1n$ for $n\ge 1$. Note that $$s_{n+1}=s_n+\frac1{(n+1)^2}<s_n+\frac1{n(n+1)}$$ (why?): this will be useful for the induction step.

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You solve my puzzle of "why subtract 1/n" :) I get it now thanks – SC L Jan 17 at 4:23
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@SCL: You’re welcome. – Brian M. Scott Jan 17 at 4:28

Hint: Instead of trying to prove that

$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} < 2,$$

try instead to prove the stronger inequality

$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} \le 2-\dfrac{1}{n}.$$

This is easily done by induction.

Hint 2: The inequality $\dfrac{1}{(n+1)^2} < \dfrac{1}{n(n+1)}$ will be helpful.

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The stronger inequality is indeed easier to prove :) I understand now thanks! – SC L Jan 17 at 4:21

$$\frac{1}{k^2}<\frac{1}{k(k-1)}$$

$$\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}=1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}<1+\frac{1}{2(2-1)}+\cdots+\frac{1}{n(n-1)}=2-\frac{1}{n}<2$$

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