ltn→∞12n+1+12n+2+12n+3+....+12n+n=

Ltx→012n+1+12n+2+.......+12n+nLtx→∞∑r=1n12n+r=Ltx→∞∑r=1n12+rn.1n=∫0112+xdx=log2+x01=log3−log2=loge3/2

ltn→∞12n+1+12n+2+12n+3+....+12n+n=

Ltx→012n+1+12n+2+.......+12n+nLtx→∞∑r=1n12n+r=Ltx→∞∑r=1n12+rn.1n=∫0112+xdx=log2+x01=log3−log2=loge3/2

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$\underset{n \to \infty}{l t} [\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + \frac{1}{2 n + 3} + .... + \frac{1}{2 n + n}] =$

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$\log_{e}^{(\frac{1}{3})}$

$\log_{e}^{(2 / 3)}$

$\log_{e}^{(3 / 2)}$

$\log_{e}^{(4 / 3)}$

answer is C.

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$\underset{x \to 0}{L t} [\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + ....... + \frac{1}{2 n + n}]$

$\underset{x \to \infty}{L t} \sum_{r = 1}^{n} \frac{1}{2 n + r} = \underset{x \to \infty}{L t} \sum_{r = 1}^{n} \frac{1}{2 + (\frac{r}{n})} . \frac{1}{n}$

$= \int_{0}^{1} \frac{1}{2 + x} d x = {[\log (2 + x)]}_{0}^{1} = \log 3 - \log 2 = \log_{e} (3 / 2)$

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ltn→∞12n+1+12n+2+12n+3+....+12n+n=

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$\underset{n \to \infty}{l t} [\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + \frac{1}{2 n + 3} + .... + \frac{1}{2 n + n}] =$