Question : Find the sum of the series 1/(n+1)+1/(n+2)+…..+1/4n as limit n approaches infinity Solution : Let the sum of the series as \[L=\lim_{n\rightarrow\infty}{\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots..+\frac{1}{4n}\right]\ }\] \[L=\lim_{n\rightarrow\infty}{\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots..+\frac{1}{n+3n}\right]\ }\] \[=\lim_{n\rightarrow\infty}{\frac{1}{n}\left[\frac{n}{n}+\frac{n}{n+1}+\frac{n}{n+2}+\ldots..+\frac{n}{n+3n}\right]\ }\] \[=\lim_{n\rightarrow\infty}{\frac{1}{n}\left[\frac{1}{1+\frac{0}{n}}+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\ldots..+\frac{1}{1+\frac{3n}{n}}\right]\ }\] \[=\lim_{n\rightarrow\infty}{\frac{1}{n}\left[\sum_{r=0}^{3n}\frac{1}{1+\frac{r}{n}}\right]\ }\] \[=\int_{0}^{3}{\frac{1}{1+x}dx}\] \[=\log{\left(4\right)}-\log{\left(1\right)}\] \[L=\log{\left(4\right)}\] Therefore \[\lim_{n\rightarrow\infty}{\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots..+\frac{1}{4n}\right]\ }=4\]
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