Question

\(\lim _{n \rightarrow \infty}\left(\sum_{r=1}^{m} r^{n}\right)^{1 / n}\) is equal to, \((n \in N)\) (A) \(\mathrm{m}\) (B) \(\mathrm{m} / 2\) (C) \(\mathrm{e}^{\mathrm{m}}\) (D) \(\mathrm{e}^{\mathrm{m} 2}\)

Short answer

Answer: (A) m

Step-by-step solution

Write down the expression

We are given the expression: $$\lim _{n \rightarrow \infty}\left(\sum_{r=1}^{m} r^{n}\right)^{1 / n}$$ where \(m\) is a fixed natural number and we want to find the equivalent option for the limit.

Express the sum as a fraction of largest term

We know that as n goes to infinity, the largest term in the sum \(\sum_{r=1}^{m} r^{n}\) will dominate the sum. So let's express each term in the sum as a fraction of the largest term \(m^n\). We can rewrite the sum as: $$\sum_{r=1}^{m} r^{n} = m^n\sum_{r=1}^{m} \left(\frac{r}{m}\right)^n$$ Now, analyze the limit again: $$\lim_{n \rightarrow \infty} \left( m^n\sum_{r=1}^{m} \left(\frac{r}{m}\right)^n \right)^{1/n}$$

Apply limit laws

We can apply the limit laws and rewrite the limit as: $$\lim_{n \rightarrow \infty} \left( m^n\sum_{r=1}^{m} \left(\frac{r}{m}\right)^n \right)^{1/n} = m\lim_{n \rightarrow \infty} \left(\sum_{r=1}^{m} \left(\frac{r}{m}\right)^n\right)^{1/n}$$

Find the dominant term

Note that in the limit we are taking, all terms in the sum inside the brackets get smaller and smaller because all \(r<m\), except for the last term when \(r=m\), which remains 1. So, $$\lim_{n \rightarrow \infty} \left(\sum_{r=1}^{m} \left(\frac{r}{m}\right)^n\right)^{1/n} = \lim_{n \rightarrow \infty} \left(1 \right)^{1/n}$$

Evaluate the limit

The limit \(\lim_{n \rightarrow \infty} \left(1 \right)^{1/n}=1\) (since 1 raised to any power is always 1). We combine this with the result from Step 3: $$m\lim_{n \rightarrow \infty} \left(\sum_{r=1}^{m} \left(\frac{r}{m}\right)^n\right)^{1/n} = m(1)$$ So, the limit of the given expression is equal to \(m\), which corresponds to the option (A).