## Start by representing the Taylor series as a power series

Sometimes we’ll be asked because that the radius and interval that convergence of a Taylor series. In stimulate to discover these things, we’ll first have to uncover a power series representation because that the Taylor series.

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Once we have the Taylor series represented together a power series, we’ll recognize ???a_n??? and also ???a_n+1??? and also plug them into the limit formula from the proportion test in order to say where the series is convergent.

## Step-by-step solution for detect the radius and interval the convergence

**Example**

Using the graph below, find the third-degree Taylor series about ???a=3??? because that ???f(x)=\ln(2x)???. Then find the power series representation of the Taylor series, and the radius and interval that convergence.

**Taylor series**

Since we currently have the graph done, the worth in the far right obelisk becomes the coefficient on each term in the Taylor polynomial, in the form

???\fracf^(n)(a)n!(x-a)^n???

With the whole chart filled in, us can develop each hatchet of the Taylor polynomial.

Putting all of the state together, we acquire the third-degree Taylor polynomial.

???\ln6+\frac13(x-3)-\frac118(x-3)^2+\frac181(x-3)^3???

**Power collection representation**

We desire to discover a power collection representation because that the Taylor series above. The very first thing we have the right to see is the the exponent of each ???(x-3)??? is same to the ???n??? value of the term, which means that

???(x-3)^n???

will be component of the power series representation. The fountain coefficient in front of the ???(x-3)??? terms deserve to be represented by

???\frac1n3^n???

Finally, we need to resolve the negative sign in front of the???n=2??? term. If we multiply our terms by

???(-1)^n+1???

the ???n=2??? term will certainly be an unfavorable and the ???n=1??? and ???n=3??? terms will be positive. Remember, nobody of this generalizations use to our ???n=0??? term, therefore we’ll leave this term outside of the power series representation.

???\ln6+\sum^\infty_n=1\frac(-1)^n+1(x-3)^nn3^n???

Notice the amount starts at ???n=1???, due to the fact that the ???n=0??? term is not included in the sum.

**Radius and also interval of convergence**

To discover the radius the convergence, we’ll recognize ???a_n??? and ???a_n+1??? making use of the power collection representation we simply found.

???a_n=\frac(-1)^n+1(x-3)^nn3^n???

???a_n+1=\frac(-1)^n+2(x-3)^n+13^n+1(n+1)???

We deserve to plug ???a_n??? and also ???a_n+1??? right into the border formula native the ratio test.

???L=\lim_n\to\infty\left|\fraca_n+1a_n\right|???

???L=\lim_n\to\infty\left|\frac\frac(-1)^n+2(x-3)^n+1(n+1)3^n+1\frac(-1)^n+1(x-3)^nn3^n\right|???

???L=\lim_n\to\infty\left|\frac(-1)^n+2(x-3)^n+1(n+1)3^n+1\cdot\fracn3^n(-1)^n+1(x-3)^n\right|???

???L=\lim_n\to\infty\left|\frac(-1)^n+2(-1)^n+1\cdot\frac(x-3)^n+1(x-3)^n\cdot\fracnn+1\cdot\frac3^n3^n+1\right|???

???L=\lim_n\to\infty\left|(-1)^n+2-(n+1)\cdot(x-3)^n+1-n\cdot\fracnn+1\cdot3^n-(n+1)\right|???

???L=\lim_n\to\infty\left|(-1)^n+2-n-1\cdot(x-3)^n+1-n\cdot3^n-n-1\cdot\fracnn+1\right|???

???L=\lim_n\to\infty\left|(-1)^1\cdot(x-3)^1\cdot3^-1\cdot\fracnn+1\right|???

???L=\lim_n\to\infty\left|-\frac13(x-3)\fracnn+1\right|???

Since we’re managing absolute value, the ???-1??? deserve to be removed.

???L=\lim_n\to\infty\left|\fracn(x-3)3(n+1)\right|???

The border only impacts ???n???, for this reason we have the right to remove the ???(x+3)???.

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???L=|x-3|\lim_n\to\infty\left|\fracn3(n+1)\right|???

???L=|x-3|\lim_n\to\infty\left|\fracn3n+3\right|???

Since we’ll get the indeterminate type ???\infty/\infty??? if we try to advice the limit, we’ll divide the numerator and denominator by the highest-degree change in order to mitigate the fraction.

???L=|x-3|\lim_n\to\infty\left|\fracn3n+3\left(\frac\frac1n\frac1n\right)\right|???

???L=|x-3|\lim_n\to\infty\left|\frac\fracnn\frac3nn+\frac3n\right|???