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.
 A: Via a comparison series and integral:
$$
\sum_{k=1}^n \frac{1}{\sqrt{k}} = 
\sum_{k=1}^n \int_{k-1}^{k}\frac{dx}{\sqrt{k}}
\leq \sum_{k=1}^n \int_{k-1}^{k}\frac{dx}{\sqrt{x}}
= \int_{0}^{n}\frac{dx}{\sqrt{x}}
= 2\sqrt{n}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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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 $\ds{\zeta\pars{1/2} < 0}$.

$$
\mbox{Also,}\quad
\sum_{k = 1}^{n - 1}{1 \over \root{k}} + {1 \over \root{n}} <
2\root{n} + \zeta\pars{1 \over 2} + {1 \over \root{n}}
$$
$$ \\
\mbox{which leads to}\quad
\bbx{\sum_{k = 1}^{n}{1 \over \root{k}} <
2\root{n + 1} + \zeta\pars{1 \over 2}}\\
$$
A: $\sum_{k=1}^n \frac{1}{\sqrt{k}}$ is a $p$-series with $p\le 1$, hence it diverges.
Alternatively,
$$0<\frac{1}{n\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt 1}=\frac1{\sqrt n}\le1.$$
A: Hint:
For $k\le x<k+1$,
$$\frac1{\sqrt k}<\frac1{\sqrt{x-1}}.$$
Integrate from $1$ to $n+1$.
