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I'll preface and apologize for the somewhat lengthy intro.

Alright, so recently I've been self-studying summation and found interest in convergent series. I was looking at some fairly common series and read that the reciprocals of triangular numbers summed indefinitely converge to $2$.

Immediately I thought of the reciprocals of powers of $2$ starting at $0$ and summed indefinitely, which also converge to $2$.

From here, I got the idea to do a proof that the two series' sums are equal. For reciprocals of powers of $2$, I didn't have any trouble getting the following:

$$\sum_{i=0}^{\infty}2^{-i} =\,\,2$$

However, writing the reciprocals of triangular numbers in summation notation was a bit more difficult. I started off with the following basic summation to simply generate triangular numbers:

$$\sum_{i=1}^n i\,\,\,|\,\,\,n\in\Bbb N_1$$ However, I needed to generate these terms some other way to function as my summand, so I used the following identity with a binomial coefficient to do so.

$$\sum_{i=1}^{n-1} i = \begin{pmatrix}n\\2\\ \end{pmatrix}$$

The difference of $1$ is negligible since we are summing indefinitely. Thus, we finally reach summation notation describing the reciprocals of triangular numbers indefinitely summed, which is the following:

$$\sum_{i=2}^{\infty} \begin{pmatrix}i\\2\\ \end{pmatrix}^{-1} =\,\,2$$

Now we get to our equation, which is:

$$ \sum_{i=0}^{\infty}2^{-i} = \sum_{i=2}^{\infty} \begin{pmatrix}i\\2\\ \end{pmatrix}^{-1}$$

This is where the proof begins. Gauss' method obviously won't work with either part since we're working with indefinite values, not to mention exponents and reciprocals. I've tried researching for infinite summation properties, but I've only found some where the summand base is less than $1$. I have no idea how to deal with a summand that's a binomial coefficient.

I'm not asking for users to do the proof for me (In fact I don't want someone to do that), but if you could give me some applicable summation identities that would help me generate a proof from here, that would be fantastic.

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  • $\begingroup$ It's not clear what you're asking. Are you asking for a proof of that last equality without first knowing that they both sum to $2$? $\endgroup$
    – Casteels
    Commented Aug 30, 2015 at 18:45
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    $\begingroup$ Note that the reciprocal of $\binom{i}{2}$ is $\frac{2}{i(i-1)}$. This is equal to $\frac{2}{i-1}-\frac{2}{i}$. Now use telescoping. $\endgroup$ Commented Aug 30, 2015 at 18:45
  • $\begingroup$ @Casteels Not asking for someone to give me a proof, just asking for help in proving that that equation is true. $\endgroup$ Commented Aug 30, 2015 at 18:58
  • $\begingroup$ @AndréNicolas Fantastic, this is precisely what I needed to solve the rightmost summation! $\endgroup$ Commented Aug 30, 2015 at 19:02
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    $\begingroup$ Good, I am pleased that given a start you were able to finish things. $\endgroup$ Commented Aug 30, 2015 at 19:09

2 Answers 2

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Hint: For $T(n)=\frac{n(n-1)}{2}$ we have $\frac{1}{T(n)}=2\left(\frac{1}{n-1}-\frac{1}{n}\right)$. And please correct the statement "the two series are equal." You mean to say the two series' sums are equal.

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  • $\begingroup$ Grammar has been fixed, thanks for the catch. Could you write that first part in $LaTeX$? $\endgroup$ Commented Aug 30, 2015 at 18:54
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    $\begingroup$ @L.A.F.2. The formatting has been done. $\endgroup$ Commented Aug 30, 2015 at 19:18
  • $\begingroup$ Fantastic, this will resolve the right side of the equation. Thanks! $\endgroup$ Commented Aug 30, 2015 at 19:20
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We have the following: $$ 2^{-i} = \Bigg(\frac{1}{2}\Bigg)^i $$

And thus: $$ \sum_{i=0}^\infty 2^{-i} =\sum_{i=0}^\infty \Bigg(\frac{1}{2}\Bigg)^i $$

Now you can use your properties for infinite sums that you have, if your base is $<1$

Infact, what you have here, is called geometric series by mathematicans. Look it up on wikipedia.

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  • $\begingroup$ Exactly what I needed. Now the left side of the equation is done. Thank you so much! $\endgroup$ Commented Aug 30, 2015 at 18:53

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