lim _{n rightarrow infty}left{frac{1}{n+m}+fr | vedclass

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(C) આપેલ પદાવલિ $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n+km}$ છે.આપણે તેને $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1

Vedclass · Accepted Answer

$\frac{\log _e(1+m)}{m}$

Vedclass · Answer

(C) આપેલ પદાવલિ $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n+km}$ છે.આપણે તેને $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1+m(\frac{k}{n})}$ તરીકે ફરીથી લખી શકીએ.નિશ્ચિત સંકલનની વ્યાખ્યા મુજબ,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n f(\frac{k}{n}) = \int_0^1 f(x) dx$.અહીં,$f(x) = \frac{1}{1+mx}$.તેથી,સંકલન $\int_0^1 \frac{1}{1+mx} dx$ બને છે.સંકલનનું મૂલ્ય: $\frac{1}{m} [\log _e(1+mx)]_0^1 = \frac{1}{m} (\log _e(1+m) - \log _e(1)) = \frac{\log _e(1+m)}{m}$.

$\lim _{n \rightarrow \infty}\left\{\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots+\frac{1}{n+n m}\right\}=$

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