lim _{n rightarrow infty} frac{1}{n^{k+1}}lef | vedclass

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(C) The given limit is $L = \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \sum_{r=1}^{n} (2r)^k$. We can rewrite this as $L = \lim _{n \rightarrow \i

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$\frac{2^k}{k+1}$

Vedclass · Answer

(C) The given limit is $L = \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \sum_{r=1}^{n} (2r)^k$. We can rewrite this as $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \left(\frac{2r}{n}\right)^k$. Using the definition of the definite integral as the limit of a Riemann sum, $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_{0}^{1} f(x) dx$. Here, $f(x) = (2x)^k$. Thus, $L = \int_{0}^{1} (2x)^k dx = 2^k \int_{0}^{1} x^k dx$. Evaluating the integral, $L = 2^k \left[ \frac{x^{k+1}}{k+1} \right]_{0}^{1} = 2^k \left( \frac{1}{k+1} - 0 \right) = \frac{2^k}{k+1}$.

$\lim _{n \rightarrow \infty} \frac{1}{n^{k+1}}\left[2^k+4^k+6^k+\ldots+(2 n)^k\right]=$

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