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Show that the following inequality holds for all integers $N\geq 1$ $$\left|\sum_{n=1}^N\frac{1}{\sqrt{n}}-2\sqrt{N}-c_1\right|\leq\frac{c_2}{\sqrt{N}}$$ where $c_1,c_2$ are some constants.

I have tried induction but it doesn't seem promising. Any ideas please?

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    $\begingroup$ Idea: Divide by $\sqrt{N}$ and get $$\left|\frac{1}{N}\sum_{n=1}^N \frac{1}{\sqrt{n/N}} -2 - \frac{c_1}{\sqrt{N}}\right|$$ The sum is a Reimann sum for $\int_{0}^1\frac{1}{\sqrt{x}}dx$. Not sure how that helps... $\endgroup$ – Thomas Andrews Jun 26 '14 at 19:52
  • $\begingroup$ Related if not duplicate $\endgroup$ – Daniel Fischer Jun 26 '14 at 19:55
  • $\begingroup$ So can you give a complete answer for this question now? Thanks $\endgroup$ – Gatsby Aug 23 '16 at 1:26
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Since $1/\sqrt{n}$ is decreasing, we can get an upper bound on the sum by computing: $$1+\int_1^N \frac{1}{\sqrt{x}}dx = 1+2\sqrt{N}$$

Thus we have $$\sum_{n=1}^N \frac{1}{\sqrt{n}} - 2\sqrt{N} = (1+2\sqrt{N}) - 2\sqrt{N} + E(N) = 1 + E(N)$$

Where $$E(N) = 1+\int_1^N \frac{1}{\sqrt{x}}dx - \sum_{n=1}^N \frac{1}{\sqrt{n}} > 0$$

is the error of our estimation at the point $N$. Now can you estimate $E(N)$?


Using the formula from Corollary 2.4 in the pdf linked in the comments, we find that $$\sum_{n=1}^N \frac{1}{\sqrt{n}} = \int_{1}^N \frac{1}{\sqrt{x}} dx - \int_{1}^N \{x\} \frac{-1/2}{(x)^{3/2}}dx + 1$$

Here $\{x\}$ is the fractional part of $x$, which is always less than 1. The middle integral can be estimated by $$\left|\int_{1}^N \{x\} \frac{-1/2}{(x)^{3/2}}dx\right| < \int_1^N 1 \cdot \frac{1/2}{(x)^{3/2}}dx = 1-\frac{1}{\sqrt{N}}$$

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  • $\begingroup$ I don't actually know if you can estimate it. To be honest. Though there are some tricks that Euler came up with that might be helpful. Also try $c_1=1$ in this case. $\endgroup$ – Joel Jun 26 '14 at 20:01
  • $\begingroup$ It has already been estimated here ($d=-1/2$) $\endgroup$ – Winther Jun 26 '14 at 20:16
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    $\begingroup$ So it is. I believe we can obtain a more straightforward estimate using Corollary 2.4 from this: math.uiuc.edu/~hildebr/ant/main2.pdf $\endgroup$ – Joel Jun 26 '14 at 20:21
  • $\begingroup$ Hi @Joel. Can you please specify what values $c_1$ and $c_2$ are? Thanks. $\endgroup$ – lovelesswang Jun 26 '14 at 21:09
  • $\begingroup$ Don't reference the comments in your answer. Just put the comment link in the answer. There might end up being many comments here, and the reader has to figure out which one you are referring to. $\endgroup$ – Thomas Andrews Jun 27 '14 at 1:03

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