Radius of Convergence of power series?
Σ [(2x+1)^n]/[(n+1)2^2n]

Question

Radius of Convergence of power series?
Σ [(2x+1)^n]/[(n+1)2^2n]

solomonzelman · Answer

$\large\color{black}{ \displaystyle \sum_{  }^{ \infty } ~ \frac{(2x+1)^n}{(n+1)(2)^{2n}}}$

like this?

solomonzelman · Answer

use the ratio test

solomonzelman · Answer

$\large\color{black}{ \displaystyle \lim_{n\rightarrow 0} \left|\left(A_{n+1}\right)\div\left(A_{n}\right)\right|}$

anonymous · Answer

Like that

anonymous · Answer

Just beed help working it out I think I know the answer but not sure

solomonzelman · Answer

oh, the limit should be limit n-> infitnty

solomonzelman · Answer

$\large\color{black}{ \displaystyle \sum_{  }^{ \infty } ~ \frac{(2x+1)^n}{(n+1)(2)^{2n}}}$

solomonzelman · Answer

$\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty} \left|\frac{(2x+1)^{n+1}}{(n+1+1)(2)^{2n+1}}\times \frac{(n+1)(2)^{2n}}{(2x+1)^{n}}\right|}$

$\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty} \left|\frac{(2x+1)^{n+1}}{(n+2)(2)^{2n+1}}\times \frac{(n+1)(2)^{2n}}{(2x+1)^{n}}\right|}$

$\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty} \left|\frac{(2x+1)}{(n+2)(2)^{2n+1}}\times \frac{(n+1)(2)^{2n}}{1}\right|}$

$\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty} \left|\frac{(2x+1)}{2(n+2)}\times \frac{(n+1)}{1}\right|}$

$\large\color{black}{ \displaystyle \left|2x+1\right| \cdot \lim_{n \rightarrow \infty} \left|\frac{n+1}{2(n+2)}\right|}$

$\large\color{black}{ \displaystyle \frac{1}{2}\left|2x+1\right| \cdot \lim_{n \rightarrow \infty} \frac{n+1}{n+2}}$

$\large\color{black}{ \displaystyle \frac{1}{2}\left|2x+1\right| \cdot 1}$

solomonzelman · Answer

So,       $\large\color{black}{ \displaystyle {\rm L}=\frac{1}{2}\left|2x+1\right| \cdot 1}$

solomonzelman · Answer

$\large\color{black}{ \displaystyle {\rm L}=\frac{1}{2}\left|2x+1\right| }$

solomonzelman · Answer

for series to converge  $\large\color{black}{ \displaystyle {\rm L}<1 }$  so...

$\large\color{black}{ \displaystyle \frac{1}{2}\left|2x+1\right|<1 }$

$\large\color{black}{ \displaystyle \left|2x+1\right|<2 }$

$\large\color{black}{ \displaystyle -2<2x+1<2 }$

$\large\color{black}{ \displaystyle -3<2x<1 }$

$\large\color{black}{ \displaystyle -\frac{3}{2}<x<\frac{1}{2} }$

solomonzelman · Answer

the interval is on you to figure out

freckles · Answer

$$\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty} \left|\frac{(2x+1)^{n+1}}{(n+1+1)(2)^{2(n+1)}}\times \frac{(n+1)(2)^{2n}}{(2x+1)^{n}}\right|}  $$ $$
\large\color{black}{ \displaystyle \frac{1}{4}\left|2x+1\right| \cdot 1}$$

freckles · Answer

$$\frac{1}{4}|2x+1|<1 $$

anonymous · Answer

The radius of convergence is 1/2 correct that is what I got, but my professor got radius of convergence is 2.

freckles · Answer

$$\frac{1}{4}|2x+1|<1 \ \frac{1}{4}(2)|x+\frac{1}{2}|<1 \ \frac{1}{2}|x+\frac{1}{2}|<1 \ |x+\frac{1}{2}|<2 $$

freckles · Answer

I have the radius is 2 there

anonymous · Answer

Yes thanks