how many terms of the convergent series sum from n = 1 to infinity of 5 over n to the power of 1.2 should be…

Explanation:

Step1: Recall the remainder formula for integral - test

For a positive - decreasing function (f(x)) such that (a_n = f(n)) and the series (\sum_{n = 1}^{\infty}a_n) converges, the remainder (R_N=\sum_{n = N+1}^{\infty}a_n\leq\int_{N}^{\infty}f(x)dx). Here, (a_n=\frac{5}{n^{1.2}}), so (f(x)=\frac{5}{x^{1.2}}).

Step2: Calculate the improper integral

We want (R_N\leq0.000001). First, calculate (\int_{N}^{\infty}\frac{5}{x^{1.2}}dx). Using the power - rule for integration (\int x^r dx=\frac{x^{r + 1}}{r+1}+C) ((r\neq - 1)), we have (\int_{N}^{\infty}\frac{5}{x^{1.2}}dx=\lim_{t\rightarrow\infty}\int_{N}^{t}5x^{-1.2}dx). [ \begin{align*} \lim_{t\rightarrow\infty}\int_{N}^{t}5x^{-1.2}dx&=\lim_{t\rightarrow\infty}\left[5\times\frac{x^{-1.2 + 1}}{-1.2+1}\right]{N}^{t}\ &=\lim{t\rightarrow\infty}\left[5\times\frac{x^{-0.2}}{-0.2}\right]{N}^{t}\ &=\lim{t\rightarrow\infty}\left[-25x^{-0.2}\right]{N}^{t}\ &=\lim{t\rightarrow\infty}\left(-\frac{25}{t^{0.2}}+\frac{25}{N^{0.2}}\right) \end{align*} ] As (t\rightarrow\infty), (\lim_{t\rightarrow\infty}\frac{-25}{t^{0.2}} = 0). So (\int_{N}^{\infty}\frac{5}{x^{1.2}}dx=\frac{25}{N^{0.2}}).

Step3: Solve for (N)

Set (\frac{25}{N^{0.2}}\leq0.000001). Cross - multiply to get (25\leq0.000001N^{0.2}). Then (N^{0.2}\geq\frac{25}{0.000001}=25000000). Raise both sides to the power of 5 (since ((N^{0.2})^5=N)): (N\geq(25000000)^5). [ \begin{align*} N&\geq(2.5\times10^{7})^5\ &=2.5^5\times10^{35}\ &=97.65625\times10^{35}\ &\approx9.77\times 10^{36} \end{align*} ]

Answer:

(9.77\times 10^{36}) (rounded up to the nearest whole number)