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Today me and my friends had a long debate on a question.

I am not writing the full question but only the main point.
Let $$f(n) = \frac{1}{S_n - \frac{1}{15}} $$

We found out $S_n$ to be: $$S_n = \sum_{k=1}^n{\frac{1}{(2n-1)(2n+1)(2n+3)(2n+5)}} $$

Question: How many roots will $f(n) = 0$ have?

I said that $f(n)$ can't be zero. So the answer will be None of these.

Because, for example:
Let $$a = x² -4x + 4$$ For $a = 0$, $x$ should be $\pm2$

But now let $$b = \frac{1}{\frac{1}{t²-4t+4} + 1} $$ Taking LCM we get: $$b = \frac{t²-4t+4}{t²-4t+5}$$ Now for $b$ to be zero, $t$ should be $\pm2$. But if t is $\pm2$ the expression $t²-4t+4$ will become zero and in initial b $1/0$ will occur which should not be there.

If $$t=\pm2$$ Then $$b = \frac{1}{\frac{1}{0} + 1}$$

So I say b can never be zero. Two of us said this and the other two were not agreeing. Am I correct or my friend?

Bill Dubuque
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Arya
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1 Answers1

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You are actually asking two different questions:

i) How many roots will $f(n)$ have?

ii) Am I correct that $b$ can never be zero?

The answer to the first is none, since for any fixed input of a real number, you do not get zero as the output.

The answer to the second one is different; in the limit as the the ratio in the denominator tends to infinity, the overall fraction tends to zero (see the link in your comments for more detailed explanation on this case). But no single value of $n$ will actually give you $f(n)=0$, so it's important to distinguish these concepts. Really what we're saying in case ii) is that we can get our output to be as close to zero as we like by taking larger and larger values as inputs, so in this sense, the limiting behavior of this process gives us zero - but we never actually get there, we just get as close as we like.

Alborz
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  • I see otherwise when it comes to the first point. Assuming the summation upper bound is $N$ not $n$ , the denominator would be an increasing function and indeed for large values of $n$, f(n) can get so close to zero, however, since its like $k/n$, it will not reach real zero. – NoChance Apr 15 '23 at 15:37
  • @NoChance not sure what you're trying to say. Getting close to zero is not the same as root, can you clarify? – Alborz Apr 15 '23 at 18:12
  • A root for f(n) makes f(n) equals zero exactly, no value of n can do this here. However, f(n) tends to be very close to zero (almost zero) for many values of n. – NoChance Apr 15 '23 at 18:13
  • @NoChance that's what I said though. So what do you see otherwise? – Alborz Apr 15 '23 at 18:14
  • I think you should mention that f(n) is almost zero because practically, its a zero whereas theoretically it is not. If this equation is for practical reasons, I would consider it to be zero. – NoChance Apr 15 '23 at 18:35
  • @NoChance A root has to make the function actually zero, not just close to zero. If you are approximating a root, then it makes sense to consider 'close-enough' cases; here, there is no root to approximate, we have an asymptotic behavior. – Alborz Apr 15 '23 at 19:25
  • Correct. However, my understanding is the OP is not looking for a "root" specifically. Anyway, I thought I'd point out what I am thinking. Thank you. – NoChance Apr 16 '23 at 03:59
  • @NoChance yep just trying to clarify. But it does say in his post specifically 'root' – Alborz Apr 16 '23 at 04:18