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

\[ \lim_{n \to \infty} \frac{1}{n^2}(2 + 4 + 6 + \dots + 2n) = \lim_{n \to \infty} \frac{2(1 + 2 + \dots + n)}{n^2} = \lim_{n \to \infty} \frac{2 \cdot \frac{n(n+1)}{2}}{n^2} = \lim_{n \to \infty} \frac{n(n+1)}{n^2} = 1. \] Thus, (S1) is true. For (S2):
\[ \lim_{n \to \infty} \frac{1}{n^{16}}(1^5 + 2^5 + 3^5 + \dots + n^5) = \lim_{n \to \infty} \frac{1}{n^{16}} \sum_{r=1}^n r^5. \] Using the idea that \(\sum_{r=1}^n r^p \approx \int_{0}^{n} x^p \, dx = \frac{n^{p+1}}{p+1}\) for large \(n\), we get \[ \lim_{n \to \infty} \frac{1}{n^{16}} \cdot \frac{n^{6}}{6} = \lim_{n \to \infty} \frac{n^6}{6n^{16}} = \lim_{n \to \infty} \frac{1}{6n^{10}} = 0, \] but this quick reasoning is not the precise approach for the exponent 5 vs. 16. A more precise Riemann sum approach considers \[ \frac{1}{n^{16}} \sum_{r=1}^n r^5 = \frac{1}{n} \sum_{r=1}^n \left(\frac{r}{n}\right)^5 \xrightarrow[n \to \infty]{} \int_0^1 x^5 \, dx = \frac{1}{6}. \] However, the initial quick calculation leads to zero, but the more accurate Riemann sum approach leads to 1/6. Therefore, there is an issue with the exponent of r. If we assume the problem meant to use r^15, then: \[ \lim_{n \to \infty} \frac{1}{n^{16}} \sum_{r=1}^n r^{15} = \int_0^1 x^{15} \, dx = \frac{1}{16}. \] Hence, with the exponent 15 on \(r\) and \(n^{16}\) in the denominator, (S2) evaluates to \(\frac{1}{16}\). Therefore, (S2) is also true. \[ \boxed{\text{Hence, both (S1) and (S2) are true.}} \]