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Exercises: Well-known Maclaurin Series

Theory

Main Idea: Generating New Series from the Geometric Series

The most important starting point is the geometric series:

\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n, \quad |x|<1

From this single formula we can:

  1. Replace xx by another expression.
  2. Differentiate both sides.
  3. Integrate both sides.

The radius of convergence changes only when we substitute a new expression inside xx.

Exercises

Exercise 1
f(x)=1x+1f(x)=\frac{1}{x+1}
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Solution

Rewrite the function so that it matches the geometric form:

1x+1=11−(−x)\frac{1}{x+1} = \frac{1}{1-(-x)}

Now use

11−x=∑n=0∞xn\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n

Replace xx by −x-x:

11−(−x)=∑n=0∞(−x)n\frac{1}{1-(-x)} = \sum_{n=0}^{\infty}(-x)^n

Simplify:

1x+1=∑n=0∞(−1)nxn\frac{1}{x+1} = \sum_{n=0}^{\infty}(-1)^n x^n

For convergence:

|-x|<1 \Rightarrow |x|<1

Conclusion:

1x+1=∑n=0∞(−1)nxn,R=1\boxed{ \frac{1}{x+1} = \sum_{n=0}^{\infty}(-1)^n x^n, \quad R=1 }
Exercise 2
f(x)=1(x+1)2f(x)=\frac{1}{(x+1)^2}
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Solution

Start from the previous result:

11+x=∑n=0∞(−1)nxn\frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^n x^n

Differentiate both sides.

Derivative of the left side:

ddx(11+x)=−1(1+x)2\frac{d}{dx}\left(\frac{1}{1+x}\right) = -\frac{1}{(1+x)^2}

Derivative of the right side:

∑n=1∞(−1)nnx n−1\sum_{n=1}^{\infty}(-1)^n n x^{\,n-1}

So,

−1(1+x)2=∑n=1∞(−1)nnx n−1-\frac{1}{(1+x)^2} = \sum_{n=1}^{\infty}(-1)^n n x^{\,n-1}

Multiply both sides by −1-1:

1(1+x)2=∑n=1∞(−1)n−1nx n−1\frac{1}{(1+x)^2} = \sum_{n=1}^{\infty}(-1)^{n-1} n x^{\,n-1}

Re-index by letting k=n−1k=n-1:

1(1+x)2=∑k=0∞(k+1)(−1)kxk\frac{1}{(1+x)^2} = \sum_{k=0}^{\infty}(k+1)(-1)^k x^k

The radius of convergence does not change under differentiation.

Conclusion:

1(1+x)2=∑n=0∞(n+1)(−1)nxn,R=1\boxed{ \frac{1}{(1+x)^2} = \sum_{n=0}^{\infty}(n+1)(-1)^n x^n, \quad R=1 }
Exercise 3
f(x)=11+4x2f(x)=\frac{1}{1+4x^2}
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Solution

Rewrite:

11+4x2=11−(−4x2)\frac{1}{1+4x^2} = \frac{1}{1-(-4x^2)}

Using the geometric series:

11−u=∑n=0∞un\frac{1}{1-u} = \sum_{n=0}^{\infty}u^n

Let u=−4x2u=-4x^2:

11+4x2=∑n=0∞(−4x2)n\frac{1}{1+4x^2} = \sum_{n=0}^{\infty}(-4x^2)^n

Simplify:

(−4x2)n=(−1)n4nx2n(-4x^2)^n = (-1)^n 4^n x^{2n}

Thus,

11+4x2=∑n=0∞(−1)n4nx2n\frac{1}{1+4x^2} = \sum_{n=0}^{\infty}(-1)^n 4^n x^{2n}

For convergence:

| -4x^2 | < 1 \Rightarrow 4x^2 < 1 \Rightarrow |x| < \frac12

Conclusion:

11+4x2=∑n=0∞(−1)n4nx2n,R=12\boxed{ \frac{1}{1+4x^2} = \sum_{n=0}^{\infty}(-1)^n 4^n x^{2n}, \quad R=\frac12 }
Exercise 4
f(x)=14+x2f(x)=\frac{1}{4+x^2}
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Solution

Factor out 44:

14+x2=14⋅11+x24\frac{1}{4+x^2} = \frac{1}{4} \cdot \frac{1}{1+\frac{x^2}{4}}

Now use

11+u=∑n=0∞(−1)nun\frac{1}{1+u} = \sum_{n=0}^{\infty}(-1)^n u^n

Let

u=x24u=\frac{x^2}{4}

Then

14+x2=14∑n=0∞(−1)n(x24)n\frac{1}{4+x^2} = \frac{1}{4} \sum_{n=0}^{\infty} (-1)^n \left(\frac{x^2}{4}\right)^n

Simplify:

=∑n=0∞(−1)n4n+1x2n= \sum_{n=0}^{\infty} \frac{(-1)^n}{4^{n+1}}x^{2n}

For convergence:

\left|\frac{x^2}{4}\right|<1 \Rightarrow |x|<2

Conclusion:

14+x2=∑n=0∞(−1)n4n+1x2n,R=2\boxed{ \frac{1}{4+x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{4^{n+1}}x^{2n}, \quad R=2 }
Exercise 5
f(x)=11−x2f(x)=\frac{1}{1-x^2}
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Solution

We start from the geometric series:

\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n, \quad |x|<1

Now notice that

11−x2=11−(x2)\frac{1}{1-x^2} = \frac{1}{1-(x^2)}

This matches the geometric form with u=x2u=x^2.

So,

11−x2=∑n=0∞(x2)n\frac{1}{1-x^2} = \sum_{n=0}^{\infty}(x^2)^n

Simplify the power:

(x2)n=x2n(x^2)^n = x^{2n}

Therefore,

11−x2=∑n=0∞x2n\frac{1}{1-x^2} = \sum_{n=0}^{\infty}x^{2n}

For convergence:

|x^2|<1

Since ∣x2∣=∣x∣2|x^2|=|x|^2, this gives

|x|<1

Conclusion:

11−x2=∑n=0∞x2n,R=1\boxed{ \frac{1}{1-x^2} = \sum_{n=0}^{\infty}x^{2n}, \quad R=1 }
Exercise 6
f(x)=ln⁡∣1+x1−x∣f(x)=\ln\left|\frac{1+x}{1-x}\right|
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Solution

We use the known Maclaurin series:

\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}x^n, \quad |x|<1

and

\ln(1-x) = -\sum_{n=1}^{\infty} \frac{1}{n}x^n, \quad |x|<1

Now use the logarithm property:

ln⁡(1+x1−x)=ln⁡(1+x)−ln⁡(1−x)\ln\left(\frac{1+x}{1-x}\right) = \ln(1+x)-\ln(1-x)

Substitute the series:

=∑n=1∞(−1)n−1nxn+∑n=1∞1nxn= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}x^n + \sum_{n=1}^{\infty} \frac{1}{n}x^n

Combine the two series:

For even powers, the terms cancel. For odd powers, the terms double.

Thus we obtain:

ln⁡(1+x1−x)=2(x+x33+x55+x77+⋯ )\ln\left(\frac{1+x}{1-x}\right) = 2\left( x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots \right)

In summation form:

ln⁡(1+x1−x)=∑n=0∞2x2n+12n+1\ln\left(\frac{1+x}{1-x}\right) = \sum_{n=0}^{\infty} \frac{2x^{2n+1}}{2n+1}

For convergence we require

|x|<1

Conclusion:

ln⁡(1+x1−x)=∑n=0∞2x2n+12n+1,R=1\boxed{ \ln\left(\frac{1+x}{1-x}\right) = \sum_{n=0}^{\infty} \frac{2x^{2n+1}}{2n+1}, \quad R=1 }