lim _{n rightarrow infty} left( frac{sqrt{1} | vedclass

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(D) આપેલ પદાવલિ $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r \sqrt{r}}{n^{5/2}}$ છે.આને $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^{3/2}

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(D) આપેલ પદાવલિ $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r \sqrt{r}}{n^{5/2}}$ છે.આને $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^{3/2}}{n^{5/2}} = \lim _{n \rightarrow \infty} \sum_{r=1}^n \left( \frac{r}{n} \right)^{3/2} \cdot \frac{1}{n}$ તરીકે ફરીથી લખી શકાય છે.સરવાળાની મર્યાદા તરીકે નિશ્ચિત સંકલનની વ્યાખ્યાનો ઉપયોગ કરતા,$\lim _{n \rightarrow \infty} \sum_{r=1}^n f\left( \frac{r}{n} \right) \frac{1}{n} = \int_0^1 f(x) dx$.અહીં,$f(x) = x^{3/2}$.આમ,લક્ષ $\int_0^1 x^{3/2} dx = \left[ \frac{x^{5/2}}{5/2} \right]_0^1 = \frac{2}{5} (1 - 0) = \frac{2}{5}$ છે.

$\lim _{n \rightarrow \infty} \left( \frac{\sqrt{1} + 2 \sqrt{2} + 3 \sqrt{3} + \ldots + n \sqrt{n}}{n^{5/2}} \right) = $

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