Sum of reciprocals of the square roots of the first N Natural Numbers

Question

$$\lim_{n\to\infty}\sum_{i=1}^n \frac{1}{\sqrt i}$$

This question was asked in an entrance test for an undergraduate program in India. I want to know how to approach such questions.

I was thinking of finding the partial sum till n and applying limit concepts to get an answer, but couldn't find such an expression. I even tried limit of a sum method to convert it into an integral but was unable to do that as well.

Answer 1

For an elementary proof of divergence, note that

$\begin{array}\\ \sqrt{i+1}-\sqrt{i} &=(\sqrt{i+1}-\sqrt{i})\dfrac{\sqrt{i+1}+\sqrt{i}}{\sqrt{i+1}+\sqrt{i}}\\ &=\dfrac{1}{\sqrt{i+1}+\sqrt{i}}\\ &<\dfrac{1}{2\sqrt{i}}\\ \end{array} $

so $\dfrac{1}{\sqrt{i}} \gt 2(\sqrt{i+1}-\sqrt{i}) $.

Therefore $\sum_{i=1}^n\dfrac{1}{\sqrt{i}} \gt \sum_{i=1}^n2(\sqrt{i+1}-\sqrt{i}) =2(\sqrt{n+1}-1) =2\sqrt{n}-2 $.

You can also get an upper bound from this.

$\begin{array}\\ \sqrt{i+1}-\sqrt{i} &=(\sqrt{i+1}-\sqrt{i})\dfrac{\sqrt{i+1}+\sqrt{i}}{\sqrt{i+1}+\sqrt{i}}\\ &=\dfrac{1}{\sqrt{i+1}+\sqrt{i}}\\ &>\dfrac{1}{2\sqrt{i+1}}\\ \end{array} $

so $\dfrac{1}{\sqrt{i+1}} \lt 2(\sqrt{i+1}-\sqrt{i}) $ or $\dfrac{1}{\sqrt{i}} \lt 2(\sqrt{i}-\sqrt{i-1}) $.

Therefore $\sum_{i=1}^n\dfrac{1}{\sqrt{i}} =1+\sum_{i=2}^n\dfrac{1}{\sqrt{i}} \lt 1+\sum_{i=2}^n2(\sqrt{i}-\sqrt{i-1}) =1+2(\sqrt{n}-1) =2\sqrt{n}-1 $.

Answer 2

The sum diverges. Its partial sums are larger than the harmonic series, which is know to diverge.

Answer 3

Simply observe that $$ \sum_{i=n}^{2n} \frac{1}{\sqrt{i}} \geq (2n - n) \frac{1}{\sqrt{2n}} = \frac{\sqrt{n}}{\sqrt{2}} \to \infty, $$ hence the series ( your sum ) diverges (is infinite).

Answer 4

Just added for your curiosity.

In his answer, marty cohen gave nice upper and lower bounds for the partial sums in your specific problem.

If we make the problem more general $(k>1)$, $$S_n^{(k)}=\sum \frac 1 {\sqrt[k] n}=H_n^{\left(\frac{1}{k}\right)}$$ where appear the generalized harmonic numbers.

Using their asymptotics $$S_n^{(k)}=\frac k{k-1} n^{(1-\frac 1k)}+\zeta \left(\frac{1}{k}\right)+\frac 1{2 n^{\frac 1k}}+O\left(\frac 1{ n^{(1+\frac 1k)}}\right)$$ where $\zeta \left(\frac{1}{k}\right)$ is a negative number which goes asymptotically to $-\frac 12$.