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I am supposed to prove that $(1/1) + (1/3) + (1/6) + \dots + (1/t_n) < 2$.

The hint is that $$ \frac2{n(n+1)} = 2\left(\frac1{n} - \frac1{n+1}\right) $$

However, I was thinking that if you take $ 2\left(\frac{1}{t_n} + \frac{1}{t_{n+1}}\right) = \frac1{t_{\frac{n}2}}$ where $n$ is even.

So my logic is you can subtract 1 from both sides multiply by 2 and you are back where you started.

eg $$ 1-1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + ... < 2 -1$$ $$ 2\left(\frac{1}{2} + \frac{1}{6} + ... \right) < 1 \times 2 $$ $$ 1 - 1 + \frac{1}{3} + ... < 2 -1$$ etc

Is this correct logic or am I missing something? Obviously by the hint this was not the proof they had in mind. So how would I prove this using the hint?

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3 Answers 3

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You're supposed to add the reciprocals of the triangular numbers, which are given by $$t_n = \frac{n(n+1)}{2} $$

Then the hint is just telling you to write in a more convenient way the reciprocal of $t_n$, so that when you add them something nice will happen. For example, try writing the first few sums using the suggested identity and try to discern a pattern from that.

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  • $\begingroup$ Ok I did what you said and it is similar to what I was trying to get at my way. I edited what I said so hopefully it makes more sense. $\endgroup$
    – qw3n
    Dec 21, 2010 at 3:29
  • $\begingroup$ @qw3n Good, I couldn't say anything else before because it could spoil the fun of solving it for yourself. Basically what you get is a telescoping series. $\endgroup$ Dec 21, 2010 at 3:38
  • $\begingroup$ Thanks for the help I figured it out. I think the first method I was using is correct it is just a little convoluted. $\endgroup$
    – qw3n
    Dec 21, 2010 at 3:59
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Your statement that $\frac{1}{t_n} + \frac{1}{t_{n+1}} = \left(\frac{1}{t_{n-1}}\right)\left(\frac{1}{2}\right)$ where n>1 is not correct. $\frac{1}{6}+\frac{1}{10}=\frac{4}{15}$ which is not equal to $\frac{1}{3}\cdot \frac {1}{2}$

It is not clear what you mean "you can always get back to the first term 1 as long as you multiply the inequality by 2."

When you say "$(1/t_4)+(1/t_5)\dots(1/t_n) < 2$. Because $(1/t_1) = 1$ and $(1/t_2) + (1/t_3) = 1$", in fact $(1/t_2) + (1/t_3) = 1/2$. If you add the three inequalities together, the right side is $3 \frac{1}{2}$ with this correction, but you were asked to prove it with $2$ on the right side.

As Adrian Barquero said, you need to follow the direction of the hint.

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  • $\begingroup$ I corrected the error the formula works when n is even. I missed that the odds do not work in the formula. $\endgroup$
    – qw3n
    Dec 21, 2010 at 3:19
  • $\begingroup$ No, $\frac{1}{21}+\frac{1}{28}=\frac{1}{12}, not \frac{1}{15}$. The general formula is $\frac{2}{n(n+1)}+\frac{2}{(n+1)(n+2)}=\frac{2(n+2)+2n}{n(n+1)(n+2)}=\frac{4}{n(n+2)}\neq\frac{2}{(n-1)n}$ except in a couple cases. $\endgroup$ Dec 21, 2010 at 3:57
  • $\begingroup$ Ok I think I have it right now I reworked all the algebra. The thing I got wrong was that the equation produces n/2 not n-1. Here is the algebra $\frac2{2n(2n+1)} + \frac2{(2n+1)(2n+2)} = \frac{n+n+1}{n(2n+1)(n+1)} = \left(\frac1{n(n+1)}\right)(2)$ Sorry, I should have been more careful with my algebra earlier. $\endgroup$
    – qw3n
    Dec 21, 2010 at 5:05
  • $\begingroup$ Now this algebra is correct, but your proof starts by assuming what you want to prove and doesn't show that the number of terms in the sum is finite. In fact, the dots at the end imply that it is infinite. What you appear to be arguing is that the limit of the sum is 2, so if you delete some terms the finite sum will be less than 2. It also appears that you just do the transformation and then undo it and get back where you started. $\endgroup$ Dec 21, 2010 at 5:32
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Using your approach, we can write the proof as follows: Choose $k$ such that $2^k+1 \geq n$. Then $$\sum_{i=1}^n \frac{2}{i(i+1)} \leq \sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}$$ Using the summation formula lets us talk about how many terms are involved and gives a name for the dummy variable. $$\sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}=1+\frac{1}{2}\sum_{i=1}^{2^{k-1}+1} \frac{2}{i(i+1)}$$ Here is where we used your relation cutting the number of terms in half. I chose the upper limit of the sum so we can continue the process. Define $$f(2^k+1) = \sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}$$ We have shown $f(2^k+1)=1+\frac{1}{2}f(2^{k-1}+1)$. We can continue the recursion to find $$f(2^k+1)=\sum_{i=0}^{k-2} 2^{-i} + \frac{1}{2^{k-1}}f(3)$$ As $f(3)=\frac{3}{2}$, we have $$f(2^k+1)=\sum_{i=0}^{k} 2^{-i} =2-\frac{1}{2^k}\lt 2$$

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  • $\begingroup$ Ok thanks a lot that makes more sense then what I was trying to do. Obviously I have a lot to learn about writing proofs. Also, if you do not mind one last question. Why do you use $2^k + 1$ instead of just $2k + 1$? The same would go for the other times you use exponents. $\endgroup$
    – qw3n
    Dec 21, 2010 at 14:44
  • $\begingroup$ Because I wanted to keep dividing by 2, using your relation, to get down to f(3). If I just used 2k+1 I could do the first step, but then k might be odd and I couldn't continue. $\endgroup$ Dec 21, 2010 at 15:09

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