# Finding limit of the sequence

Given a sequence defined as

$${a_{n}}= \frac{1} {\sqrt{2}} + \frac{2} {3} + \frac{3} {\sqrt{28}} + \cdots \cdots + \frac{n} {\sqrt{n^3+1}}$$

and I want to find the limit of this sequence.

I think the limit is positive infinity, but I don't know what I should do in order to prove this.

Please give me some hints on how to approach this, thanks to anybody who helps.

$$\frac{n}{\sqrt{n^3+1}}\sim\frac1{\sqrt{n}},$$ hence $$\lim_{n\to\infty}a_n=\infty.$$

Explanation, using series' methods

$$\lim_{n\to\infty}\frac{\dfrac{n}{\sqrt{n^3+1}}}{\dfrac1n}=1$$ and the series $$\lim_{n\to\infty}\frac1{\sqrt{n}}$$ is divergent, hence also $$\lim_{n\to\infty}\frac{n}{\sqrt{n^3+1}}$$ id divergent, but $a_n$'s are partial sums of this series.

More elementary explanation

It is easy to see, that $$\dfrac{n}{\sqrt{n^3+1}}\geq\frac1n$$ for all $n\geq2$. Let $$b_n=1+\frac12+\dots+\frac1n.$$ It is certainly increasing, hence it remains to show, that it is unbounded. Let us consider $$b_{2^n}=1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\dots+\frac18\right)+\dots +\left( \frac1{2^{n-1}+1}+\dots+\frac1{2^n}\right).$$ But the value in every parentheses is greater than $\frac12$, hence $$b^{2^n}>1+\frac{n}{2}.$$

• Sorry, but could you please explain this in a bit more detail？
– Lucy
Nov 2, 2014 at 2:28
• @Lucy Certainly, but I don't know, which tools are allowed. Do you know something about series or it is an initial course of calculus? Nov 2, 2014 at 2:33
• This is for my analysis lesson, and I have learnt about sum rule, product rule, quotient rule, and ratio tests. I am also allowed to use some standard limit results.
– Lucy
Nov 2, 2014 at 2:41
• @Lucy OK, hence the explanation by series, which I have put, is a bit to advanced. Please wait for an alternatibve version. Nov 2, 2014 at 2:47

Note that for $n > 1$

$$\frac{n}{\sqrt{n^3 +1}} = \frac{1}{\sqrt{n + 1/n^2 }}> \frac1{\sqrt{2n}}> \frac1{\sqrt{2}n},$$

and the series diverges because the harmonic series $\sum 1/n$ diverges.

• Sorry, but I didn't learn about the comparison test yet, could you please explain using another method?
– Lucy
Nov 2, 2014 at 2:59
• Do you know that $\sum 1/n$ diverges to infinity? You have $a_n > (1/\sqrt{2})\sum_{k=1}^{n} 1/k$.
– RRL
Nov 2, 2014 at 3:04