tutorial exercise evaluate the indefinite integral as a power series. what is the radius of convergence r…

tutorial exercise evaluate the indefinite integral as a power series. what is the radius of convergence r? ∫t/(1 - t^12) dt step 1 using 1/(1 - r)=∑(n = 0)^∞ r^n, a power series representation of 1/(1 - t^12) is ∑(n = 0)^∞(t^12)^n. step 2 since 1/(1 - t^12)=∑(n = 0)^∞(t^12n), a power series representation of t/(1 - t^12)=t(1/(1 - t^12)) is ∑(n = 0)^∞(t^12n + 1). step 3 now, ∫t/(1 - t^12) dt = c+∑(n = 0)^∞( ) submit skip (you cannot come back)

tutorial exercise evaluate the indefinite integral as a power series. what is the radius of convergence r? ∫t/(1 - t^12) dt step 1 using 1/(1 - r)=∑(n = 0)^∞ r^n, a power series representation of 1/(1 - t^12) is ∑(n = 0)^∞(t^12)^n. step 2 since 1/(1 - t^12)=∑(n = 0)^∞(t^12n), a power series representation of t/(1 - t^12)=t(1/(1 - t^12)) is ∑(n = 0)^∞(t^12n + 1). step 3 now, ∫t/(1 - t^12) dt = c+∑(n = 0)^∞( ) submit skip (you cannot come back)

Answer

Explanation:

Step1: Recall geometric - series formula

We know that $\frac{1}{1 - r}=\sum_{n = 0}^{\infty}r^{n}$, for $|r|\lt1$. Let $r = t^{12}$, then $\frac{1}{1 - t^{12}}=\sum_{n = 0}^{\infty}(t^{12})^{n}=\sum_{n = 0}^{\infty}t^{12n}$.

Step2: Multiply by $t$

Since $\frac{1}{1 - t^{12}}=\sum_{n = 0}^{\infty}t^{12n}$, then $\frac{t}{1 - t^{12}}=t\times\frac{1}{1 - t^{12}}=\sum_{n = 0}^{\infty}t\times t^{12n}=\sum_{n = 0}^{\infty}t^{12n + 1}$.

Step3: Integrate term - by - term

We know that $\int\frac{t}{1 - t^{12}}dt=\int\left(\sum_{n = 0}^{\infty}t^{12n+1}\right)dt$. Integrating term - by - term, $\int t^{12n + 1}dt=\frac{t^{12n+2}}{12n + 2}+C$. So $\int\frac{t}{1 - t^{12}}dt=C+\sum_{n = 0}^{\infty}\frac{t^{12n + 2}}{12n+2}$.

Answer:

$C+\sum_{n = 0}^{\infty}\frac{t^{12n + 2}}{12n+2}$