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

Question

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.

Answer 1

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

Answer 2

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

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

Answer 3

Hint:

For $k\le x<k+1$,

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

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

Answer 4

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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, $$ {1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x < {1 \over 2}\int_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}} $$

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

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$$ \mbox{Also,}\quad \sum_{k = 1}^{n - 1}{1 \over \root{k}} + {1 \over \root{n}} < 2\root{n} + \zeta\pars{1 \over 2} + {1 \over \root{n}} $$ $$ \\ \mbox{which leads to}\quad \bbx{\sum_{k = 1}^{n}{1 \over \root{k}} < 2\root{n + 1} + \zeta\pars{1 \over 2}}\\ $$