Answer to Question #346725 in Real Analysis for Nikhil rawat

Question #346725

Check whether the sequence (an), where


an = 1/ (n+1) + 1/(n+2) +....+1/(2n) is convergent or not


1
Expert's answer
2022-06-06T14:35:11-0400

Consider


"1+\\dfrac{1}{2}+...+\\dfrac{1}{n}+\\dfrac{1}{n+1}+...+\\dfrac{1}{2n}"

"=(1+\\dfrac{1}{2}+...+\\dfrac{1}{n})+(\\dfrac{1}{n+1}+...+\\dfrac{1}{2n})"

Then


"\\dfrac{1}{n+1}+...+\\dfrac{1}{2n}=(1+\\dfrac{1}{2}+\\dfrac{1}{3}+...+\\dfrac{1}{n}"

"+\\dfrac{1}{n+1}+...+\\dfrac{1}{2n-1}+\\dfrac{1}{2n})-2(\\dfrac{1}{2}+\\dfrac{1}{4}+...+\\dfrac{1}{2n})"

"=1-\\dfrac{1}{2}+\\dfrac{1}{3}-\\dfrac{1}{4}+...+\\dfrac{1}{2n-1}-\\dfrac{1}{2n}"

"=\\displaystyle\\sum_{k=1}^{2n}\\dfrac{(-1)^{k+1}}{k}"

Consider "\\ln(1+x)" for "x\\in(-1, x]"


"\\ln(1+x)=\\displaystyle\\sum_{k=1}^{\\infin}\\dfrac{(-1)^{k+1}x^k}{k}"

Then for "x=1"


"\\ln2=\\ln(1+1)=\\displaystyle\\sum_{k=1}^{\\infin}\\dfrac{(-1)^{k+1}(1)^k}{k}"

"=\\displaystyle\\sum_{k=1}^{\\infin}\\dfrac{(-1)^{k+1}}{k}"

We see that


"\\lim\\limits_{n\\to\\infin}a_n=\\lim\\limits_{n\\to\\infin}(\\dfrac{1}{n+1}+...+\\dfrac{1}{2n})"

"=\\lim\\limits_{n\\to\\infin}\\displaystyle\\sum_{k=1}^{2n}\\dfrac{(-1)^{k+1}}{k}=\\displaystyle\\sum_{k=1}^{\\infin}\\dfrac{(-1)^{k+1}}{k}=\\ln2"

Therefore he sequence "(a_n)" is convergent.


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