Question

In Exercises \(35-40,\) find a formula for the sum of \(n\) terms. Use the formula to find the limit as \(n \rightarrow \infty\). $$ \lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{16 i}{n^{2}} $$

Short answer

The formula for the sum of n terms is \( \frac{8(n+1)}{n} \) and the limit as \( n \rightarrow \infty \) is 8.

Step-by-step solution

Find the formula for the sum of n terms

Given is the formula from i=1 to n for \( \frac{16 i}{n^{2}} \). To find the sum of n terms, find the constant factor and the variable part. You can recognize that \( 16/n^2 \) is a constant and \( i \) is the variable part which changes with each term. This can be written as \( (16/n^2) * ∑_{i=1}^{n} i \)

Calculate the Sum of n terms

∑_{i=1}^{n} i can be found using the formula: ∑_{i=1}^{n} i = \( \frac{n(n+1)}{2} \). Substituting it back into the equation, get \( (16/n^2) * \frac{n(n+1)}{2} = \frac{16(n+1)}{2n} = \frac{8(n+1)}{n} \). This is the formula for the sum of n terms.

Find the Limit as n tends to Infinity

To find the limit as \( n \rightarrow \infty \), substitute \( \infty \) for n in the formula from step 2: \( \lim _{n \rightarrow \infty} \frac{8(n+1)}{n} \). Apply the rule for finding the limit of a fraction, that says when the degree of the numerator and the denominator is the same, the limit is the ratio of the leading coefficients. In this case, the limit is 8.