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$$ \lim_{n \longrightarrow \infty} \frac{1}{\sqrt1}+ \frac{1}{\sqrt 2}+...+\frac{1}{\sqrt n} $$

I have been stuck with this limit for some time and need some help to keep going. I can see it won't converge but I obviously have to prove it. I tried to rewrite it as:

$$ \lim_{n \longrightarrow \infty} \frac{1+\sqrt 2}{\sqrt 2}+ \frac{\sqrt 3 + \sqrt 4}{\sqrt {12}}+...+\frac{\sqrt n + \sqrt {n-1}}{\sqrt {n(n-1)}}, $$

but it wasn't very useful.

I also tried to approach it as a Riemann sum, but I didn't really see the light either because I couldn't figure out anything more than

$$ \lim _{n \longrightarrow \infty}\sum ^n _{i=1} \frac{1}{\sqrt i} $$

I would really appreciate any help.

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    $\begingroup$ This diverges. Indeed $\sum \frac 1n$ diverges and this is term by term larger than that. $\endgroup$
    – lulu
    Commented Dec 31, 2021 at 12:47
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    $\begingroup$ The Riemann sum way: $$\sum_{i=1}^n\frac1{\sqrt i}=\sqrt n\:\cdot\left(\frac1n\sum_{i=1}^n\frac1{\sqrt{\frac in}}\right)\!.$$ $\endgroup$
    – nejimban
    Commented Dec 31, 2021 at 12:57

2 Answers 2

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The sum has $n$ terms and each term is bounded from below by $\frac{1}{\sqrt{n}}$, so the sum is bounded from below by $\frac{n}{\sqrt{n}} = \sqrt{n}$. Of course, $\lim_{n \to \infty} \sqrt{n} = +\infty$.

(Using integrals or sophisticated criteria is an overkill in this case.)

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Here is another way to approach it differently from the comparison test.

More precisely, you can apply the condensation test.

If $a_{n}$ is non-negative and decreasing, then the following relation holds: \begin{align*} \sum_{n=1}^{\infty}a_{n} \ \ \text{converges} \ \ \text{iff} \ \ \sum_{n=1}^{\infty}2^{n}a_{2^{n}} \ \ \text{converges} \end{align*}

Generically speaking, the following series \begin{align*} \sum_{n=1}^{\infty}\frac{1}{n^{\alpha}} = 1 + \frac{1}{2^{\alpha}} + \frac{1}{3^{\alpha}} + \ldots \end{align*} converges iff the following series \begin{align*} \sum_{n=1}^{\infty}2^{n}\times\frac{1}{(2^{n})^{\alpha}} = \sum_{n=1}^{\infty}[2^{(1 - \alpha)}]^{n} \end{align*}

converges, which is a geometric series with ratio $r = 2^{1 - \alpha}$.

As it is known, a geometric series converges iff its ratio satisfies $|r| < 1$.

Consequently, the proposed series converges iff \begin{align*} 2^{1 - \alpha} < 1 & \Longleftrightarrow 2^{1 - \alpha} < 2^{0}\\\\ & \Longleftrightarrow 1 - \alpha < 0\\\\ & \Longleftrightarrow \alpha > 1 \end{align*}

and diverges otherwise. Hence we conclude the proposed series diverges since $\alpha = 1/2$.

Hopefully this helps !

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