Functions Question 314
Question: If $ S_{n}=\sum\limits_{k=1}^{n}{a_{k}} $ and $ \underset{n\to \infty }{\mathop{\lim }},a_{n}=a, $ then $ \underset{n\to \infty }{\mathop{\lim }},\frac{{S_{n+1}}-S_{n}}{\sqrt{\sum\limits_{k=1}^{n}{k}}} $ is equal to
Options:
A) 0
B) a
C) $ \sqrt{2}a $
D) $ 2a $
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ \underset{n\to \infty }{\mathop{\lim }}\frac{{S_{n+1}}-S_{n}}{\sqrt{\sum\limits_{k=1}^{n}{k}}}=\underset{n\to \infty }{\mathop{\lim }}\frac{{a_{n+1}}}{\sqrt{\frac{n,(n+1)}{2}}}=0 $ (Since $ n\to \infty ,,\text{numerator }\to a\text{ while}\text{denominator }\to \infty \text{)} $
BETA
