Question

\((S1) : lim_{n→∞}\frac{1}{ n ^2} ( 2 + 4 + 6 + . . . . + 2n) = 1\)
\((S2):  lim_n→∞\frac{1}{n^{16}}( 1^15 +2^15 +3^15 + . . . . + n^15 ) = \frac{1}{16} \)

• A. Only (S1) is true
• B. Only (S2) is true
• C. Both (S1) and (S2) are true
• D. Both (S1) and (S2) are false

Answer 1

The Correct Option is C

To determine the truth of the statements \(S1\) and \(S2\), let's analyze each one separately using mathematical reasoning and limit calculations.

Statement \((S1): \lim_{n→∞}\frac{1}{ n^2} ( 2 + 4 + 6 + \cdots + 2n) = 1\)

The series \(2 + 4 + 6 + \cdots + 2n\) is an arithmetic series where each term is \(2 \times k\), with \(k\) ranging from 1 to \(n\). Thus, it can be rewritten as:

2(1 + 2 + 3 + \cdots + n)

The sum inside the parentheses is the sum of the first \(n\) natural numbers, which equals \(\frac{n(n+1)}{2}\). Therefore, we have:

2 \times \frac{n(n+1)}{2} = n(n+1)

Plugging this into the limit expression, we get:

\lim_{n→∞}\frac{n(n+1)}{n^2} = \lim_{n→∞}\left(\frac{n^2 + n}{n^2}\right) = \lim_{n→∞}\left(1 + \frac{1}{n}\right) = 1

Thus, Statement \(S1\) is true.

Statement \((S2): \lim_{n→∞}\frac{1}{n^{16}}( 1^{15} +2^{15} +3^{15} + \cdots + n^{15} ) = \frac{1}{16} \)

The series in the limit is the sum of the 15th powers of the first \(n\) natural numbers. In such sums, the largest term (i.e., \(n^{15}\)) dominates as \(n\) approaches infinity. Hence, we approximate:

1^{15} + 2^{15} + 3^{15} + \cdots + n^{15} \approx \frac{n^{16}}{16} \text{ (by degree analysis and polynomial approximation of sums)}

Substitute this into the limit expression:

\lim_{n→∞}\frac{1}{n^{16}} \times \frac{n^{16}}{16} = \frac{1}{16}

This shows that Statement \(S2\) is also true.

Therefore, the correct answer is: Both \((S1)\) and \((S2)\) are true.