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Let $n$ be a positive natural number, $n\ge 2$. Then $\displaystyle\sum_{k=1}^n \frac{1}{k^2} \lt 2 - \frac{1}{n}.$

The basis step was easy but could someone give me a hint in the right direction as to how to do the induction step?

I tried this:

$\displaystyle\sum_{k=1}^k \frac{1}{k^2} + \frac{1}{(k + 1)^2} \lt 2 - \frac{1}{k + 1}$

But it's getting me nowhere or I am doing something wrong. I am no expert so a clear explanation would be appreciated. Thanks.

Martin Sleziak
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1337holiday
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    Perhaps this might also help in a general sort of way: http://math.stackexchange.com/questions/19485/dominoes-and-induction-or-how-does-induction-work/19488#19488 – Arturo Magidin Apr 19 '11 at 01:59
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    See http://math.stackexchange.com/questions/1251544/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 25 '15 at 07:07

2 Answers2

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Here is how the induction step should look:

$$ \text{Assume } \sum_{k = 1}^{n} \frac{1}{k^2} < 2 - \frac{1}{n}.$$

Then,

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

Now the problem is reduced to showing that

$$ - \frac{1}{n} + \frac{1}{(n+1)^2} \leq - \frac{1}{n+1} $$

which is easy to show with some algebra. The point is that you have to use the assumption that it works for $n$. Also, when you use $k$ as an index over which you are summing, you should not use $k$ anywhere else like you did above.

Hope this helps.

Bill Karr
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  • I noticed you added the 1/(n + 1)^2 to both sides, can you explain why this was done? I think thats what i was missing. – 1337holiday Apr 19 '11 at 01:42
  • I pulled the $n+1$ term off of the sum. We assumed that what was left was less than $2 - \frac{1}{n}$. The $n+1$ term still remains. – Bill Karr Apr 19 '11 at 01:48
  • That is how induction works... you assume the formula or statement is true for some $n$, and then you try to show the statement is true for $n+1$. – Bill Karr Apr 19 '11 at 01:49
  • The heart of induction is that first you prove a base case (here n=2), then you prove that if it is true for n then it is true for n+1. So he is proving it true for n+1, given that it is true for n. Look at the upper limit of the sums-that is where the $\frac{1}{(n+1)^2} comes from. – Ross Millikan Apr 19 '11 at 01:52
  • What about the summation, how did you use the induction hypothesis to get rid of that? – 1337holiday Apr 19 '11 at 01:52
  • @1337holiday: The induction hypothesis says the summation from 1 through n is less than $2 - \frac{1}{n}$. That's how he applied the induction (he did not "get rid of" the sum, he compared it with something we were assuming was larger). – Arturo Magidin Apr 19 '11 at 01:58
  • @Arturo im still confused as to where to go next or how he reduced the problem. Im not getting this, if you could please elaborate a bit more. – 1337holiday Apr 19 '11 at 02:14
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    @1337holiday: Say your wallet has 10 bills in it, and you want to show that you have less than 100 dollars in it; if you know that the last bill is either a 20 or less; and you know that the first 9 bills add up to at most 60 dollars, you don't need to count all the bills: you can just say: "the first 9 are at most 60, and the last bill is at most 20, so the ten bills together add up to at most 80 dollars, less than 100." Here you have a sum of $n+1$ things. And you know (thanks to the induction hypothesis) that the first $n$ add up to at most $2 - \frac{1}{n}$; (cont) – Arturo Magidin Apr 19 '11 at 02:18
  • @1337holiday: (cont)... That means that all $n+1$ things add up to at most $2-\frac{1}{n}$ plus whatever the "last" term is. That's the last step in the second displayed equation. – Arturo Magidin Apr 19 '11 at 02:19
  • @Arturo That was well put. I completely understand what you are trying to say. My problem however is using that hypothesis to do the algebra part of it. How can i use that assumption to proceed with the question? Can you put me one step further? – 1337holiday Apr 19 '11 at 02:32
  • @1337holiday: I beg to differ, but your last question tells me that you did not "completely understand" what I was saying, because if you had, you would not have asked that question. The induction hypothesis is used precisely when we write that$$\left(\sum_{k=1}^n\frac{1}{k^2}\right)+\frac{1}{(n+1)^2} \lt \left(2 - \frac{1}{n}\right) + \frac{1}{(n+1)^2}.$$ That's the assumption, that the first summand on the left is less than the first summand on the right. That's it. You don't use it for any "algebra", you use it for that inequality. – Arturo Magidin Apr 19 '11 at 02:42
  • So if we know that, this is like knowing you have the $60 in your wallet right (for the first summand)? And the 1/(n + 1)^2 is the last bill that was given? – 1337holiday Apr 19 '11 at 02:52
  • @1337holiday: Assuming/knowing this inequality is like assuming/knowing that the first 9 bills of your wallet add up to at most 60 (that's the "induction hypothesis"). The extra summand is "plus whatever's left" (the last bill). – Arturo Magidin Apr 19 '11 at 02:55
  • So what im thinking then i since we already know the first summand is less than the first summand of the RHS. We just need to show that if we add the extra summand to both sides that the LHS is still less than the RHS? – 1337holiday Apr 19 '11 at 03:02
  • @1337holiday: You are trying to show that $$\sum_{k=1}^{n+1}\frac{1}{k^2} \lt 2 - \frac{1}{n+1}.$$ You know the first $n$ summands on the left hand side are at most $2 - \frac{1}{n}$, so you know that the entire left hand side is at most $$\left(2 - \frac{1}{n}\right) + \frac{1}{(n+1)^2}.$$ This is still not what you want. Now what you need to show is that $$\left( 2 - \frac{1}{n}\right) + \frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}.$$ You already know that the left hand side of this inequality is greater than what you have, but you are trying to show what you have is less than the RHS – Arturo Magidin Apr 19 '11 at 03:36
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What you write is actually what you are trying to show. However, you are almost there. For the induction step, you get to assume that

$$ \sum_{k=1}^n \frac{1}{k^2} < 2 - \frac{1}{n}. $$

Then, you need to show the statement holds for $k = n+1$:

$$ \sum_{k=1}^{n+1} \frac{1}{k^2} = \sum_{k=1}^n \frac{1}{k^2} + \frac{1}{(n+1)^2} < \dots $$

JavaMan
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