lim(n tending to infinity) (1/(n+1)+1/(n+2)+.....+1/6n)=
the sum of the above series is
i have done by graph and answer is coming out to be ln6
how should i solve theorotically
$\displaystyle \displaystyle\lim_{n \to{+}\infty}{\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+ \ldots+\dfrac{1}{6n}\right)}=$
$\displaystyle \displaystyle\lim_{n \to{+}\infty}{\dfrac{1}{n}}\sum_{k=1}^{5n}\dfrac{1 }{1+k/n}=\int_0^5\dfrac{dx}{1+x}=\log 6$