Limit of a sum $\frac{1}{\sqrt n}$ [duplicate]

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

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.

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

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.)

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 !