The sum of the reciprocal of the triangular numbers up to $\frac1{t_n}$ is < 2

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $$f(2^k+1)=\sum_{i=0}^{k} 2^{-i} =2-\frac{1}{2^k}\lt 2$$