Sequence And Series Question 69
Question: The sum to infinity of the given series $ \frac{1}{n}-\frac{1}{2n^{2}}+\frac{1}{3n^{3}}-\frac{1}{4n^{4}}+…. $ is
[MP PET 1994]
Options:
A) $ {\log_{e}}( \frac{n+1}{n} ) $
B) $ {\log_{e}}( \frac{n}{n+1} ) $
C) $ {\log_{e}}( \frac{n-1}{n} ) $
D) $ {\log_{e}}( \frac{n}{n-1} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{1}{n}-\frac{1}{2n^{2}}+\frac{1}{3n^{3}}-\frac{1}{4n^{4}}+…. $ $ =\frac{1}{n}-\frac{{{(1/n)}^{2}}}{2}+\frac{{{(1/n)}^{3}}}{3}-\frac{{{(1/n)}^{4}}}{4}+…. $ $ ={\log_{e}}( 1+\frac{1}{n} )={\log_{e}}( \frac{n+1}{n} ) $ .
BETA
