Limit problem in book: $\lim_{n\to\infty}\frac{1+\sqrt{2}+…+\sqrt{n}}{n^{3/2}}$

My book states that the $$\lim_{n\to\infty}\frac{1+\sqrt{2}+...+\sqrt{n}}{n^{3/2}}= \lim_{n\to\infty}\frac{1}{n}\left[\frac{1+\sqrt{2}+...+\sqrt{n}}{n^{1/2}}\right]=\frac{2}{3}.$$ But I just don't see it. I was thinking that if the $$\lim_{n\to\infty}\frac{1}{n}=0$$ then the whole thing would equal 0 but I don't see how to get $\frac{2}{3}$. Thanks

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Interpret it as a Riemann or Darboux sum for $\sqrt{x}$ over $[0,1]$. –  LutzL Feb 22 '14 at 5:46
yes that is what it is but how do they get that the limit is $\frac{2}{3}$? I know how to get it by calculating the integral of $\sqrt{x}$ from $[0,1]$. –  user60887 Feb 22 '14 at 5:47
The limit of the Riemann sums, if it exists, i.e., if the function is Riemann integrable, is just the value of the integral. –  LutzL Feb 22 '14 at 5:55
So basically just calculate the integral? I was trying to find the limit value just just like that if I did not know if it was the riemann sum of $\sqrt{X}$ on $[0,1]$. –  user60887 Feb 22 '14 at 5:59

3 Answers

Draw the graph of $y=\sqrt{x}$. In red, draw the rectangle with base $[0,1]$ and height $\sqrt{1}$, the rectangle with base $[1,2]$ and height $\sqrt{2}$, and so on up to the rectangle with base $[n-1,n]$ and height $\sqrt{n}$.

In blue, draw the rectangle with base $[1,2]$ and height $\sqrt{1}$, the rectangle with base $[2,3]$ and height $\sqrt{2}$, and so on up to the rectangle with base $[n,n+1]$ and height $\sqrt{n}$.

Then $\sqrt{1}+\sqrt{2}+\cdots +\sqrt{n}$ is the sum of the areas of the red rectangles, and it is also the sum of the areas of the blue rectangles. Call this sum $S_n$.

Then $S_n$ is greater than the area under $y=\sqrt{x}$ from $x=0$ to $x=n$, and less than the area under $y=\sqrt{x}$ from $x=1$ to $x=n$. Thus $$\int_0^n \sqrt{x}\,dx\lt S_n\lt \int_1^{n+1} \sqrt{x}\,dx.$$ Calculate. We get $$\frac{2}{3}n^{3/2} \lt S_n\lt \frac{2}{3} ((n+1)^{3/2}-1).$$ Divide through by $n^{3/2}$. We get $$\frac{2}{3}\lt \frac{S_n}{n^{3/2}}\lt \frac{2}{3}\frac{(n+1)^{3/2}-1}{n^{3/2}}.$$ Now let $n\to\infty$. The expression $\frac{(n+1)^{3/2}-1}{n^{3/2}}$ has limit $1$, so by Squeezing we get $\lim_{n\to\infty} \frac{S_n}{n^{3/2}}=\frac{2}{3}$.

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Another variant is to use the theorem of Cesaro-Stolz. There

\begin{align} \lim_{n\to\infty}\frac{1+\sqrt2+...+\sqrt n}{n^{3/2}} &=\lim_{n\to\infty}\frac{\sqrt{n}}{n^{3/2}-(n-1)^{3/2}} \\ &=\lim_{n\to\infty}\frac{\sqrt{n}(n^{3/2}+(n-1)^{3/2})}{n^3-(n-1)^3} \\ &=\lim_{n\to\infty}\frac{n^2(1+(1-\tfrac1n)^{3/2})}{3n^2-3n+1}=\frac23 \end{align}

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As $\displaystyle\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$

$$\lim_{n\to\infty}\frac1n\sum_{1\le r\le n}\sqrt{\frac rn}=\int_0^1\sqrt xdx$$

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