# Upper bound of $\sum_{k=1}^n \frac{1}{\sqrt{k}}$?

I am looking for an upper bound of $$\sum_{k=1}^n \frac{1}{\sqrt{k}}$$. Alternatively, is the sequence $$\frac{1}{n\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt{k}}$$ bounded?

I am trying to use a Strong law of Large Numbers by Feller and need to show this condition.

• Maybe specify a bit about what you are looking for would help. Say, $k$ is a straightforward upper bound for the sum... Commented Jul 4, 2020 at 18:48
• Note: $$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\leq\int_{0}^{n}\frac{\mathrm{d}x}{\sqrt{x}}=2\sqrt{n}.$$ Commented Jul 4, 2020 at 18:49
• Not in terms of $O(1)$ but the bound is of form $\sum_{n=1}^{k}\frac{1}{\sqrt{n}}=O(\sqrt{k})$. Commented Jul 4, 2020 at 18:50
• Thanks! That's what I needed. Commented Jul 4, 2020 at 18:55
• I meant $n{}{}$ Commented Jul 5, 2020 at 1:00

Via a comparison series and integral: $$\sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \int_{k-1}^{k}\frac{dx}{\sqrt{k}} \leq \sum_{k=1}^n \int_{k-1}^{k}\frac{dx}{\sqrt{x}} = \int_{0}^{n}\frac{dx}{\sqrt{x}} = 2\sqrt{n}$$

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Lets use a powerful Riemann Zeta Function Identity:

\begin{align} \sum_{k = 1}^{n}{1 \over \root{k}} & = \sum_{k = 1}^{n}{1 \over k^{\color{red}{1/2}}} = {n^{1 - \color{red}{1/2}} \over 1 - \color{red}{1/2}} + \zeta\pars{\color{red}{1 \over 2}} + \color{red}{1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{\color{red}{1 /2} +1}}\,\dd x \\[5mm] & = 2\root{n} + \zeta\pars{1 \over 2} + {1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x \end{align} However, $${1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x < {1 \over 2}\int_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}}$$

$$\\ \mbox{Then,}\quad \bbx{\sum_{k = 1}^{n}{1 \over \root{k}} < 2\root{n} + \zeta\pars{1 \over 2} + {1 \over \root{n}}}\\$$ Note that $$\ds{\zeta\pars{1/2} < 0}$$.
$$\mbox{Also,}\quad \sum_{k = 1}^{n - 1}{1 \over \root{k}} + {1 \over \root{n}} < 2\root{n} + \zeta\pars{1 \over 2} + {1 \over \root{n}}$$ $$\\ \mbox{which leads to}\quad \bbx{\sum_{k = 1}^{n}{1 \over \root{k}} < 2\root{n + 1} + \zeta\pars{1 \over 2}}\\$$

$$\sum_{k=1}^n \frac{1}{\sqrt{k}}$$ is a $$p$$-series with $$p\le 1$$, hence it diverges.

Alternatively, $$0<\frac{1}{n\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt 1}=\frac1{\sqrt n}\le1.$$

• Thanks! Indeed, I know it diverges. I needed an upper bound on the rate at which it diverges, and learned that an upper bound on the rate is $\sqrt{n}$. I am a bit confused by your proof of the upper bound of the sequence. Can you please explain? Commented Jul 4, 2020 at 19:44

Hint:

For $$k\le x,

$$\frac1{\sqrt k}<\frac1{\sqrt{x-1}}.$$

Integrate from $$1$$ to $$n+1$$.