If {S_n} = sumlimits_{k = 1}^n {{a_k}} and m | vedclass

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(A) We are given that ${S_n} = \sum\limits_{k = 1}^n {{a_k}}$.Therefore,${S_{n + 1}} - {S_n} = {a_{n + 1}}$.We know that $\sum\limits_{k = 1}^n k = \f

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(A) We are given that ${S_n} = \sum\limits_{k = 1}^n {{a_k}}$.Therefore,${S_{n + 1}} - {S_n} = {a_{n + 1}}$.We know that $\sum\limits_{k = 1}^n k = \frac{n(n + 1)}{2}$.Thus,the limit becomes $\mathop {\lim }\limits_{n 	o \infty } \frac{{{a_{n + 1}}}}{{\sqrt{\frac{n(n + 1)}{2}}}} = \mathop {\lim }\limits_{n 	o \infty } \frac{{{a_{n + 1}} \cdot \sqrt{2}}}{{\sqrt{n(n + 1)}}}$.Since $\mathop {\lim }\limits_{n 	o \infty } {a_n} = a$,we have $\mathop {\lim }\limits_{n 	o \infty } {a_{n + 1}} = a$.As $n 	o \infty$,the numerator approaches $a \cdot \sqrt{2}$ (a finite value) and the denominator $\sqrt{n(n + 1)} 	o \infty$.Therefore,the limit is $\frac{a \cdot \sqrt{2}}{\infty} = 0$.

If ${S_n} = \sum\limits_{k = 1}^n {{a_k}} $ and $\mathop {\lim }\limits_{n \to \infty } {a_n} = a,$ then $\mathop {\lim }\limits_{n \to \infty } \frac{{{S_{n + 1}} - {S_n}}}{{\sqrt {\sum\limits_{k = 1}^n k } }}$ is equal to

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