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$$ \lim_{n\to \infty} {\sum_{k=1}^n {1\over\sqrt{k}}\over\sqrt{n}} $$

I understand that the summation is divergent and this is $\infty\over\infty$ form. But how to proceed further??

zhw.
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princethewinner
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5 Answers5

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Using Stolz-Cesaro theorem (which is pretty useful) we get:

$\displaystyle\lim_{n\to \infty} {\sum_{k=1}^n {1\over\sqrt{k}}\over\sqrt{n}}=\lim_{n\to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=2$

Poppy
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$$\lim_{n\to \infty}\frac{\sum_{k=1}^{n}\frac{1}{\sqrt k}}{\sqrt n}=\lim_{n\to\infty}\frac 1n\sum_{k=1}^{n}\frac{1}{\sqrt{\frac kn}}=\int_{0}^{1}\frac{1}{\sqrt x}\ dx=2.$$

mathlove
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  • @mathlove, that is not always true for improper integrals. We can find a positive unbounded $f$ on $(0,1]$ such that $\int_0^1f$ converges (in the improper Riemann integral sense), yet $\lim_{n\to\infty}\frac 1n\sum_{n=1}^{k}f(\frac kn)$ doesn't exist. – zhw. Apr 06 '15 at 19:07
  • @zhw.: You are right. Thank you for pointing it out. I deleted my misleading comment. – mathlove Apr 06 '15 at 19:31
2

Use the the inequalities: $$ \sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\leq\frac{1}{2\sqrt n} \leq \frac{1}{\sqrt{n}+\sqrt{n-1}}=\sqrt{n}-\sqrt{n-1}$$

and as a consequence: $$2\frac{\sqrt{n+1}-1}{\sqrt n}\leq {\sum_{k=1}^n {1\over\sqrt{k}}\over\sqrt{n}}\leq 2$$

and the limit is $2$

Elaqqad
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1

I'll throw this solution in the mix: Let $S_n$ be the sum in the numerator. Using crayons and rectangles, we see$$\int_1^n \frac{dx}{\sqrt x} < S_n < 1+ \int_1^n \frac{dx}{\sqrt x}.$$The integral equals $2\sqrt n - 2.$ Divide by $\sqrt n$ and the limit of $2$ follow easily.

zhw.
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0

Hint: You can use L'Hopital's rule. So the limit becomes

$$\lim_{n \to \infty} \frac{2 \sqrt{n} }{\sqrt{n}} =2.$$

science
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