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In computing a sum, my book writes $$\sum_{7\leq k^2\leq 45} \frac{1}{k} = \sum_{\lceil \sqrt{7} \rceil \leq k \leq \lfloor \sqrt{45}\rfloor} \frac{1}{k}$$ How did they do this? I'm not looking for a proof that the two sums are indeed equal, I'm looking for a general rule that justifies the equality and can be applied to other similar situations.

user236343
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  • It's not true if we allow $k$ to be negative. Regardless, they're simply solving for $k$ in the original series of inequalities, nothing special – PrincessEev Mar 10 '22 at 01:13
  • @EeveeTrainer Where did the floors and ceilings come from? – user236343 Mar 10 '22 at 01:18
  • It is because the series is implicitly iterated over integer valued $k$. @user236343 They are technically not required, but useful as a reminder . – Graham Kemp Mar 10 '22 at 01:32
  • @GrahamKemp Well there are other integers besides ceil(sqrt(7)) and floor(sqrt(45)). Why did they pick these bounds instead of e.g. ceil(sqrt(7)) and the result of rounding sqrt(45) to the nearest integer? – user236343 Mar 10 '22 at 01:36

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As Eevee Trainer points out in the comments, this isn't correct in general. In fact, if we allow $k$ to be any integer, the left-hand sum is $0,$ while the right-hand sum is positive. However, I assume that in context, $k$ is known to be non-negative.

As for why they're equivalent if $k\geq 0,$ note that for such integers $k,$ we have that the following are equivalent:

  • $7\leq k^2$
  • $\sqrt 7\leq k$
  • $\left\lceil\sqrt{7}\right\rceil\leq k$

The following are also equivalent to each other for integers $k\geq 0$:

  • $k^2\leq 45$
  • $k\leq\sqrt 45$
  • $k\leq\left\lfloor\sqrt{45}\right\rfloor$
Cameron Buie
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