Question

The value of \(\lim\limits_{n \rightarrow \infin}\left(1+\frac{1}{2^n}+\frac{1}{3^n}+...+\frac{1}{(2023)^n}\right)^\frac{1}{n}\) is equal to _____________ . (rounded off to two decimal places)

Answer 1

Correct Answer: 0.99 - 1.01

To solve the limit problem, we evaluate: \(\lim\limits_{n \rightarrow \infty}\left(1+\frac{1}{2^n}+\frac{1}{3^n}+\dots+\frac{1}{2023^n}\right)^\frac{1}{n}\).
The expression inside the limit is a geometric-like series. As \(n\) approaches infinity, terms like \(\frac{1}{k^n}\) for \(k \geq 2\) tend towards zero because \(k^n\) grows exponentially.
 

Step-by-step Solution:

1. Analyze each term: \(\frac{1}{k^n}\) shrinks rapidly to zero for any integer \(k \geq 2\) as \(n \to \infty\).
2. Focus on the dominant term inside the series, which is \(1\); since other terms become negligible:

\(\left(1+\frac{1}{2^n}+\frac{1}{3^n}+\dots+\frac{1}{2023^n}\right) \approx 1\) as \(n \to \infty\).
 

1. Evaluate the \(n^{th}\) root: \(\left(1\right)^{\frac{1}{n}} = 1\).

Thus, \(\lim\limits_{n \rightarrow \infty}\left(1+\frac{1}{2^n}+\frac{1}{3^n}+\dots+\frac{1}{2023^n}\right)^\frac{1}{n} = 1\).
This value, \(1.00\), is firmly within the given range [0.99, 1.01].