Şimdi sorunuzu cevaplayabiliriz.
\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\ln\sqrt[n]{1+\frac{k}{n}}
=
\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\ln\left({1+\frac{k}{n}}\right)^{\frac{1}{n}}
=
\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{1}{n}\ln\left({1+\frac{k}{n}}\right)
=
\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{2-1}{n}\ln\left({1+k\frac{2-1}{n}}\right)
=
\lim_{n\rightarrow \infty}\frac{2-1}{n}\sum_{k=1}^{n}\ln\left({1+k\frac{2-1}{n}}\right)
=
\int_{1}^{2}\ln xdx
=
(x\ln x-x)_{1}^{2}
=
2\ln 2-2-(0-1)
=
\ln 4-\ln e
=
\ln\left(\frac{4}{e}\right)
Not: f:[a,b]\rightarrow \mathbb{R} Riemann integrallenebilir bir fonksiyon olmak üzere
\lim_{n\rightarrow\infty}\frac{b-a}{n}\sum_{k=1}^{n}f\left({a+k\frac{b-a}{n}}\right)=\int_{a}^{b}f(x)dx