4
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$$\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{121}\cdots \lt\frac{1}{4}$$ This is an interesting summation in which the addition of the next term must make the sum $\lt\frac{1}{4}$. All terms must be the reciprocal of squares $\gt2$. The first term really wouldn't be 1/4 because $\frac{1}{4}\not\lt\frac{1}{4}$. After the term $\frac{1}{36}$, it wouldnt't be $\frac{1}{49}$ because of the before-discussed rules and that 1/49 would make the sum (again) $\not\lt$ 1/4. Now the question:

If $f(1)=\frac{1}{9}$

$f(2)=\frac{1}{16}$

$f(3)=\frac{1}{25}$

$f(4)=\frac{1}{36}$

$f(5)=\frac{1}{121}$

and so on...

Is there a function that applies to this through all possible terms? In other words, does f(n)=nth term a feasible function?

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  • 1
    $\begingroup$ I see .. interesting question! (+1) $\endgroup$ – r9m Jul 9 '15 at 0:21
  • $\begingroup$ Can someone enlighten me why people think this sequence, along with the $\frac{1}{p} < 1$ sequence posted recently, qualifies as interesting? As far as I'm concerned, the sequence itself is not interesting since it's a completely arbitrary requirement (why $\frac{1}{4}$? why not $1$? or $\frac{1}{2}$? or $\frac{\pi^2}{12}$?). It would be interesting if the sequence formed had some relation to another sequence somehow so we could ask "why is there this relation?" but otherwise I find the question to be extremely boring and arbitrary. $\endgroup$ – MCT Jul 9 '15 at 0:39
  • $\begingroup$ I actually think it's a cool idea. $\endgroup$ – ncmathsadist Jul 9 '15 at 0:44
  • $\begingroup$ @ncmathsadist Maybe the premise of the sequence is "cool", but until someone gives me a reason to believe that anything interesting would come out of this (e.g similarity to another sequence or some theory), I just don't see what we gain out of studying a sequence so arbitrarily defined. $\endgroup$ – MCT Jul 9 '15 at 0:56
2
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Given a partial sum $s$ whose last term is $\dfrac{1}{n^2}$, the next term is $\dfrac{1}{m^2}$ where $m=\max(n+1,\bigg\lceil \dfrac{1}{\sqrt{\frac{1}{4}-s}}\bigg\rceil)$.

The first few terms correspond to these $n$:

3
4
5
6
11
54
519
59429
22852059
244010721780

You need multiple precision to compute this. $59429$ is as far as you can go with standard double precision floating point.

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0
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So, essentially, you have a decreasing positive sequence $(a_n)_{n=1}^{\infty}$ such that $\sum_{n=1}^{\infty} a_n$ converges and $a_m < \sum_{n=m+1}^{\infty} a_n $ for all $n$.

Let $s(m) =\sum_{n=1}^{m} a_n $.

Then you want to show that, for $r > 0$, $s(m) < r < s(m+1)$ for some $m$, there is a subsequence $(a_{n_j})_{j=1}^{\infty}$ with $n_j < n_{j+1}$ such that $r =\sum_{j=1}^{\infty} a_{n_j} $.

This would seem to follow from the fact that $\sum_{n=1}^{\infty} a_n$ converges and the growth restriction on the $a_n$.

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