$\lim_{n \to \infty}\frac{\sum_{i=0}^{n}{\sqrt{i}}}{n^2}=0 \; ?$

Question

I was wondering whether this limit converges to zero: $$ \lim_{n \to \infty}\frac{\sum_{i=0}^{n}{\sqrt{i}}}{n^2}=0 $$ And i'm pretty sure it is.

First, by intuition. I know that $\sum_{i=0}^n{i} = \frac{n(n+1)}{2}$ ~ $O(n^2)$, so i guess that $\sum_{i=0}^n{\sqrt{i}}$ is "less powerful", but i don't really know how much "lesser"

So, the thing that really interest me was: what is the "cardinality" of $\sum_{i=0}^{\infty}{\sqrt{i}} \;? \quad$ Is it "equal ~" to $O(n)$? (I'm not sure i translated the words correctly. Does 'cardinality' is the right word for my description? I'm not familiar with these words in english, sorry. hope you understood what i meant).

Here is my thought:

$ {\sum_{i=0}^{n}{\sqrt{i}}} = {\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n-1}+\sqrt{n}} = {\sqrt{n} \big( \frac{\sqrt{1}}{\sqrt{n}} +\frac{\sqrt{2}}{\sqrt{2}} +...+ \frac{\sqrt{n-1}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{n}} \big)} = {{\sqrt{n} \big( \sqrt{\frac{1}{n}} + \sqrt{\frac{2}{n}} +...+ \sqrt{\frac{n-1}{n}} + \sqrt{\frac{n}{n}} \big)}} $

and so:

$$ \lim_{n \to \infty} \frac{\sum_{i=0}^{n}{\sqrt{i}}}{n^2} = \lim_{n \to \infty} \frac{{\sqrt{n} \big( \sqrt{\frac{1}{n}} + \sqrt{\frac{2}{n}} +...+ \sqrt{\frac{n-1}{n}} + \sqrt{\frac{n}{n}} \big)}}{n^2} \overbrace{<}^{(*)} \lim_{n \to \infty} \frac{{\sqrt{n} \big( \sqrt{\frac{n}{n}} + \sqrt{\frac{n}{n}} +...+ \sqrt{\frac{n}{n}} + \sqrt{\frac{n}{n}} \big)}}{n^2} = \lim_{n \to \infty} \frac{{\sqrt{n} \big( \overbrace{1 + 1 +...+ 1 + 1}^{n \, times} \big)}}{n^2} = \lim_{n \to \infty} \frac{ \sqrt{n} *n }{n^2} = \lim_{n \to \infty} \frac{ \sqrt{n} }{n} = \lim_{n \to \infty} \frac{1}{ \sqrt{n} } = 0 $$

But my enlargement in $(*)$ above was too big.

I was wondering if you can suggest me some better way :)

Answer 1

We don't say "this limit converges to zero". We either say, "the limit of a sequence is zero", or "a sequence converges to zero".

"Cardinality" is a concept for sets; it does not make sense to say the "cardinality of a series". You may be looking for the phrase "growth rate" for a sequence.

Now back to your question. The sequence in the question can be written as $$ b_n=\frac{a_n}{\sqrt{n}}\quad \textrm{where } a_n=\left(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots+\sqrt{\frac{n}{n}}\right)\frac1n\; $$

But $a_n$ can be taken as a Riemann sum and $$ \lim_{n\to\infty}a_n=\int_0^1\sqrt{x}dx $$

It follows that $$ \lim_{n\to\infty}\frac{a_n}{\sqrt{n}}=\lim_{n\to\infty}a_n\lim_{n\to\infty}\frac{1}{\sqrt{n}}=0\quad $$

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As Matt E. pointed out in the comment, your attempt actually solves the problem! Notice that $$ |a_n|\le (\underbrace{1+1+\cdots+1}_{n\textrm{ terms}})\cdot \frac{1}{n}=1 $$ shows that $a_n$ is a bounded sequence. On the other hand you have $\lim \frac{1}{\sqrt{n}}=0$. Together you have $\lim b_n=0$.

Answer 2

You can make very good approximations using generalized harmonic numbers $$\sum_{i=0}^n \sqrt i=H_n^{\left(-\frac{1}{2}\right)}$$ For large values of $n$, then $$\sum_{i=0}^n \sqrt i=\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta \left(-\frac{1}{2}\right)+O\left(\frac{1}{n^{3/2}}\right)$$ $$\frac 1{n^2}\sum_{i=0}^n \sqrt i=\frac{2 }{3\sqrt n}+\frac{1 }{2n\sqrt n}+\cdots$$

Use it for $n=100$; the trucated series gives $\frac{403}{6000}=0.0671667$ while the exact value is $0.0671463$ (relative error of $0.03$%)

Answer 3

Integrals can be a very useful tool to get equivalent of sequences. Notice that we have for each integer $k \geq 2$ :

$$\int_{k}^{k+1} \sqrt{t} \,dt \geq {\sqrt{k}} \geq \int_{k-1}^{k} \sqrt{t} \,dt$$ since the square root function is an increasing function.

If we sum these inequalities, we get :

$$ \int_{1}^{n+1} \sqrt{t} \,dt \geq \sum_{k=1}^n \sqrt{k} \geq 1+\int_{1}^{n} \sqrt{t} \,dt $$

Then by integrating : $$ \frac{1}{1.5}(n+1)^{1.5}-\frac{1}{1.5} \geq \sum_{k=1}^n \sqrt{k} \geq \frac{1}{1.5} n^{1.5}-\frac{1}{1.5}+1 $$ Therefore, we can deduce (the thing that interests you) $$ \sum_{k=1}^n \sqrt{k} \sim_{+\infty} \frac{n^{1.5}}{1.5} $$

So we get the same result as you :

$$ \lim _{n \rightarrow \infty} \frac{\sum_{i=0}^{n} \sqrt{i}}{n^{2}} = \lim _{n \rightarrow \infty} \frac{n^{1.5}}{1.5n^{2}} = \lim \frac{1}{1.5\sqrt{n}} = 0 $$

If you are interested, using the same method we can obtain something more general $$ \sum_{k=1}^{n} \frac{1}{k^{\alpha}} \sim_{+\infty} \frac{n^{1-\alpha}}{1-\alpha}$$

for any $\alpha<1$ and

$$ \sum_{k=1}^{n} \frac{1}{k} \sim_{+\infty} \ln n$$ for the special case $\alpha=1$.