lim ⁡ 𝑛 → ∞ ∑ 𝑖

It looks like you're dealing with a sum and a limit as $n \to \infty$. Let me clarify and restate the expression you provided:

$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n} \left( 1 + \frac{2i}{n} \right)^{10}$

This type of limit involving a sum is often handled by converting it into an integral as $n \to \infty$, where the sum resembles a Riemann sum. Here's how we can approach this:

Step 1: Recognize the Riemann Sum

The given sum looks like a Riemann sum, where:

• $\frac{1}{n}$ corresponds to $\Delta x$ (the width of small intervals).
• $1 + \frac{2i}{n}$ is a function evaluated at points $\frac{i}{n}$, which tends to $x$ as $n \to \infty$.

Thus, we can rewrite the sum as an approximation of an integral. The limits of the sum, $i = 1$ to $n$, correspond to the interval $x = 1$ to $x = 3$ (since $1 + \frac{2i}{n}$ ranges from 1 to 3 as $i$ goes from 1 to $n$).

Step 2: Write the Integral

The Riemann sum approximates the following integral as $n \to \infty$:

$\int_{1}^{3} \left( x \right)^{10} \, dx$

Step 3: Solve the Integral

Now, compute the integral:

$\int_{1}^{3} x^{10} \, dx = \left[ \frac{x^{11}}{11} \right]_{1}^{3}$

Evaluating the integral:

= \frac{3^{11} - 1}{11}$$ ### Final Answer: Thus, the value of the original limit is: $$\boxed{\frac{3^{11} - 1}{11}}$$ Would you like a detailed explanation of any step, or do you have any questions? Here are five related questions to deepen your understanding: 1. How do Riemann sums relate to definite integrals? 2. What are the general steps to converting a sum to an integral? 3. How does the behavior of functions change as $$n \to \infty$$? 4. Can all sums involving limits be converted into integrals? 5. How does the power of 10 in $$(1 + \frac{2i}{n})^{10}$$ affect the limit? **Tip:** Converting sums to integrals is a powerful method in calculus, particularly when dealing with limits that grow large!