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?
Answer
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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