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  4. displaystyle limn→∞ ( (1/3.7 ) + (1/7.11) + (1/11.15) + ... + (n textterms)) =

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Q. $\displaystyle\lim_{n\to\infty} \left( \frac{1}{3.7 } + \frac{1}{7.11} + \frac{1}{11.15} + ... + \left(n\, \text{terms}\right)\right) = $

AP EAMCETAP EAMCET 2019

Solution:

Given,
$ \displaystyle\lim _{n \rightarrow \infty}\left(\frac{1}{3 \cdot 7}+\frac{1}{7 \cdot 11}+\frac{1}{11 \cdot 15}+\ldots+(n \text { terms })\right) $
$= \displaystyle\lim _{n \rightarrow \infty} \frac{1}{4}\left(\frac{4}{3 \cdot 7}+\frac{4}{7 \cdot 11}+\frac{4}{11 \cdot 15}+\ldots\right) $
$=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15} \ldots\right.\\ \left.-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+4}\right) $
$ = \displaystyle\lim _{n \rightarrow \infty} \frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{12}-\displaystyle\lim _{n \rightarrow \infty} \frac{1}{4(n+4)} $
$= \frac{1}{12}-\displaystyle\lim _{n \rightarrow \infty} \frac{1}{4 n\left(1+\frac{4}{n}\right)}=\frac{1}{12}-0=\frac{1}{12} $