find the sum sum((cos (frac(n m)/(3^n)))(3^n))

To find the sum \( \sum \left(\frac{\cos \left(\frac{n m}{3^{n}}\right)}{}\right) \), we need to evaluate the expression for each term in the series and then add them up.<br /><br />Let's consider the general term in the series: \( \frac{\cos \left(\frac{n m}{3^{n}}\right)}{} \).<br /><br />The cosine function oscillates between -1 and 1. Therefore, the maximum value of \( \cos \left(\frac{n m}{3^{n}}\right) \) is 1, and the minimum value is -1.<br /><br />To find the sum, we need to consider the behavior of the cosine function as \( n \) increases. As \( n \) becomes large, the argument of the cosine function \( \frac{n m}{3^{n}} \) approaches 0. This is because the denominator \( 3^{n} \) grows much faster than the numerator \( n m \).<br /><br />As the argument of the cosine function approaches 0, the cosine function approaches 1. Therefore, the general term \( \frac{\cos \left(\frac{n m}{3^{n}}\right)}{} \) approaches 1 as \( n \) becomes large.<br /><br />Hence, the sum \( \sum \left(\frac{\cos \left(\frac{n m}{3^{n}}\right)}{}\right) \) approaches \( n \) as \( n \) becomes large.