Q 29. - Calculus

Question

Find the interval of convergence for power series: $\sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(x-\pi \right)}^k$

Step-by-Step Solution

Answer

The interval of convergence for power series is $[-1+\pi ,1+\pi )$.

Step 1. Given information.

The given power series is $\sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(x-\pi \right)}^k$.

Step 2. Find the interval of convergence.

Let us assume $b_k=\frac{2k+1}{k^3}{\left(x-\pi \right)}^k$ therefore $b_{k+1}=\frac{2\left(k+1\right)+1}{{\left(k+1\right)}^3}{\left(x-\pi \right)}^{k+1}$

Ratio for the absolute convergence is 

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{2(k+1)+1}{{\left(k+1\right)}^3}}{\left(x-\pi \right)}^{k+1}}{{\displaystyle \frac{2k+1}{k^3}}{\left(x-\pi \right)}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{2k+3}{{\left(k+1\right)}^3}\left(x-\pi \right)\frac{k^3}{2k+1}\right| \\ =\underset{k\to \infty }{\lim }\left|x-\pi \right|{\left(\frac{k}{k+1}\right)}^3\frac{2k+3}{2k+1} \end{gathered}$$

Here the limit is $\left|x-\pi \right|$ So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\left|x-\pi \right|<1$ that is $-1+\mathrm{\pi }<\mathrm{x}<1+\mathrm{\pi }$.

Step 3. Find the interval of convergence.

Now, since the intervals are finite so we analyse the behavior of the series at the endpoints 

So, when   $x=-1+\mathrm{\pi }$

$$\begin{gathered} \sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(x-\mathrm{\pi }\right)}^k=\sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(-1+\mathrm{\pi }-\mathrm{\pi }\right)}^k \\ =\sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(-1\right)}^k \end{gathered}$$

The result is the alternating multiple of the harmonic series, which converges conditionally. 

So, when $x=1+\pi$

$$\begin{gathered} \sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(x-\mathrm{\pi }\right)}^k=\sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(1+\mathrm{\pi }-\mathrm{\pi }\right)}^k \\ =\sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(1\right)}^k \\ =\sum _{k=1}^{\infty }\frac{2k+1}{k^3} \end{gathered}$$
The result is just a constant multiple of the harmonic series which diverges 

Therefore, the interval of convergence of the power series is $\left(-1+\mathrm{\pi },1+\mathrm{\pi }\right)$.