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I have to solve this sequence:

$$a_n = \frac{1+\sqrt2+\sqrt 3+ \cdots +\sqrt n}{n\sqrt n}$$

As a tip I've been told to use Stoltz-Cesaro for sequance of this form : $a_n = \dfrac{x_n}{y_n}$

So I did Stoltz-Cesaro $\dfrac{x_n - x_{n-1}}{y_n-y_{n-1}}$ and I end up with: $\dfrac{\sqrt{n}}{n\sqrt{n}-(n-1)\sqrt{n-1}}$. I am stuck at this point, can you please give me some tips on what to do next? Thank you.

Martin Sleziak
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user137209
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  • You can do $\frac{\sqrt{n}}{n\sqrt{n}-(n-1)\sqrt{n-1}}=\frac{\sqrt{n}(n^{3/2}+(n-1)^{3/2})}{(n^{3/2}-(n-1)^{3/2})(n^{3/2}+(n-1)^{3/2})}=\frac{\sqrt{n}(n^{3/2}+(n-1)^{3/2})}{(n^{3}-(n-1)^{3})}=\frac{n^{1/2}(n^{3/2}+(n-1)^{3/2})}{-3n^2+3n-1}$. – OR. Mar 30 '14 at 15:28
  • I see, it works just fine this way. Thank you very much ABC! – user137209 Mar 30 '14 at 15:40

2 Answers2

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This is just a Riemann sum:

$$ \frac 1n \sum_{k=1}^n \sqrt{\frac kn}\to \int_0^1 x^{1/2} dx =\frac 23[x^{3/2}]_0^1 =\frac 23. $$

mookid
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A possible approach:

$$\frac{\sqrt n}{n\sqrt n-(n-1)\sqrt{n-1}}=\frac1{n-(n-1)\sqrt{1-\frac1n}}=$$

$$=\frac{n+(n-1)\sqrt{1-\frac1n}}{n^2-(n-1)^2\left(1-\frac1n\right)}=\frac{n+(n-1)\sqrt{1-\frac1n}}{n^2-n^2+n+2n-2-1+\frac1n}=$$

$$\frac{\left(1+\sqrt{1-\frac1n}\right)n-\sqrt{1-\frac1n}}{3n-3+\frac1n}\xrightarrow[n\to\infty]{}\frac23$$

DonAntonio
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