How to fing a limit $$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{kn}}?$$ I had only two minds about this. First of them was that it looks like $\frac{1^k +…+n^k}{n^{k+1}}$ which limit is $\frac{1}{k+1}$. The second is to look at square of $1+\frac{1}{\sqrt{2}} + .. + \frac{1}{\sqrt{n}}$, but it is a kind of a monster. I think, I need a formula for $1+\frac{1}{\sqrt{2}} + .. + \frac{1}{\sqrt{n}}$, but I don’t know it.
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1You know that you can take the $\frac{1}{\sqrt{n}}$ in front of the sum, right? – Matti P. Nov 06 '18 at 06:56
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Yes, of course. But I don’t know what to do after that. – Vremennik Nov 06 '18 at 06:57
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@Vremennik You should show your work and effort on it. Do not ask questions without context to avoid downvoting and closure. – user Nov 06 '18 at 06:59
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@Vremennik You need to add that effort and try directly in the question not in comments. – user Nov 06 '18 at 07:14
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@gimusi I added it in the question. – Vremennik Nov 06 '18 at 07:29
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@Vremennik Well done, what about the given hint? – user Nov 06 '18 at 07:34
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See also: What is the value of $\lim\limits_{n\to \infty}\left(\sum\limits_{i=1}^n{ \frac{1}{\sqrt{i \cdot n}} } \right)$?, How to evaluate the sum $\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2n}}+\cdots+\frac{1}{\sqrt{n^2}}$ when $n$ grows?. – Martin Sleziak May 10 '20 at 13:13
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1Does this answer your question? How to evaluate the sum $\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2n}}+\cdots+\frac{1}{\sqrt{n^2}}$ when $n$ grows? – Keen-ameteur May 10 '20 at 15:00
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HINT
We have that
$$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{kn}}=\lim_{n \to \infty} \frac1n\sum_{k=1}^n \frac{1}{\sqrt{\frac k n}}$$
then refer to
user
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1Here the function $f(x) =1/\sqrt{x}$ is not Riemann integrable on $[0,1]$ as it is unbounded. But the integral on $[0,1]$ exists as an improper Riemann integral and due to monotone nature of function the Riemann sum approach can be used here. See this answer for link between Riemann sum and improper Riemann integral: https://math.stackexchange.com/a/1745876/72031 – Paramanand Singh Nov 06 '18 at 10:13
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If you manage to prove that $$ H_n^{(1/2)}=\sum_{k=1}^{n}\frac{1}{\sqrt{k}} = 2\sqrt{n}(1+o(1)) \tag{1}$$ you are done, the limit is $2$. On the other hand
$$ \frac{H_{n+1}^{(1/2)}-H_n^{(1/2)}}{2\sqrt{n+1}-2\sqrt{n}}=\frac{2\sqrt{n+1}+2\sqrt{n}}{4\sqrt{n+1}}\to 1\tag{2}$$
hence $(1)$ is a straightforward consequence of Cesàro-Stolz.
Through creative telescoping or just induction you may actually prove the stronger bound
$$2\sqrt{n}-\tfrac{3}{2}\leq H_n^{(1/2)}\leq 2\sqrt{n}-1.\tag{3}$$
Jack D'Aurizio
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