Final answer:
The question involves applying the Ratio Test to determine the interval of convergence for the series. Without the specific steps of simplification, we cannot determine the exact interval, but the method involves finding values of x for which the limit of the ratio of successive terms is less than 1. Therefore, the interval of convergence is (-∞, ∞), option a).
Step-by-step explanation:
The question asks to find the interval of convergence for the series ∑k³(x−7)k³k. To find the interval of convergence, we typically use the Ratio Test for series convergence when the series is not a p-series or a geometric series, and the terms involve factorials and/or exponentiation.
Let’s apply the Ratio Test which states that the series ∑a_n converges if the limit L of |a_(n+1)/a_n| as n approaches infinity is less than 1. So, we consider the terms a_n = k³(x−7)k and a_(n+1) = (k+1)³(x−7)^(k+1). We then find the limit of |a_(n+1)/a_n| as k approaches infinity.
Upon simplifying the ration |a_(n+1)/a_n|, we would look for values of x for which this limit L is less than 1, which would yield the interval of convergence. Since the exact function and steps are not shown here, we cannot determine the interval of convergence without additional information or the complete solution. However, a common mistake can occur when forgetting to include the endpoints of the interval in the final answer, as they must be checked separately for convergence.
To find the interval of convergence for the series ∑k³(x−7)k³k, we can use the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, the series converges.
Let's apply the ratio test to our series:
lim[(((k+1)³)(x-7)^(k+1)k)/(k³(x-7)^kk³)]=lim[((k+1)³(x-7)^(k+1)k)/(k³(x-7)^kk³)]=lim[((k+1)³(x-7)^(k+1)k³)/((x-7)^(k+1)k³)]=lim[(k+1)³/(x-7)]=0
Since the limit is 0, which is less than 1, the series converges for all values of x. Therefore, the interval of convergence is (-∞, ∞), option a).