Infinite Series and Power Series

A ball is dropped from a height of \(1\) yard and rebounds to \(\frac{2}{3}\) of its previous height on each bounce. Find the total distance it travels.

Solution.

The initial drop covers \(1\) yard and is counted once. After that, the ball rises to height \(\frac23\) and falls back through the same height, then rises to \(\frac49\) and falls back, and so on. Every bounce height is therefore traversed twice, which is where the factor of \(2\) comes from:

The sum is geometric with first term \(a = \frac23\) and ratio \(r = \frac23\). Since \(|r| < 1\) it converges, and by \(\eqref{eq:geometric-infinite}\)

Therefore

Note that the ball bounces infinitely many times yet travels only a finite distance — infinitely many terms can still add up to a finite total.

This is a geometric series with \(a = \frac{3}{10},\ r = \frac{1}{10}\):

The repeating part is geometric with:

• \(a = \frac{285714}{10^6}\)

• \(r = \frac{1}{10^6}\)

Thus:

A beam of light of intensity \(I_0\) strikes a glass plate. At each surface a fraction \(R\) of the intensity is reflected and a fraction \(T = 1 - R\) is transmitted. Find the total intensity that emerges from the far side, accounting for light that bounces back and forth inside the plate.

Solution.

Follow the beam through the plate:

• It crosses the front surface, keeping a fraction \(T\), then crosses the back surface, keeping another \(T\). Emerging intensity: \(T^2 I_0\).

• Alternatively it reflects off the back surface (\(R\)), reflects off the front surface (\(R\)), and then exits. Each such round trip costs a factor \(R^2\).

So the emerging beams have intensities

which is geometric with first term \(a = T^2 I_0\) and ratio \(r = R^2\). Because \(0 \le R < 1\) we have \(|r| < 1\), so by \(\eqref{eq:geometric-infinite}\)

As a check: if \(R = 0\) (no reflection) then \(I_{\text{total}} = I_0\), as it must be. Note that the naive answer \(T^2 I_0\) is too small — the multiple reflections matter.

Evaluate:

Solution.

The highest power of \(n\) is \(n^4\) in both the numerator and the denominator (note that \(\sqrt{9n^8} = 3n^4\)), so divide both by \(n^4\). Inside the square root this means dividing by \(n^8\):

Now every term of the form \(1/n^k\) tends to \(0\) as \(n \to \infty\), leaving

Evaluate:

Solution: Using L'Hôpital's Rule

Note: Although \(n\) is discrete, the limit is evaluated using a continuous extension \(f(x) = \frac{\ln x}{x}\).

Evaluate:

Solution.

Take logarithm:

Hence:

Test the series

Solution.

The general term is \(a_n = \dfrac{n}{n+1}\). Dividing numerator and denominator by \(n\) gives \(a_n = \dfrac{1}{1 + 1/n} \to 1\) as \(n \to \infty\). Since this limit is \(1 \neq 0\), the series diverges by the Preliminary Test.

Test \(\displaystyle\sum_{n=1}^\infty \frac{1}{n!}\) for convergence.

Solution.

We compare against a geometric series. Both \(n!\) and \(2^n\) are products of \(n\) factors, but from the fourth factor onward every factor of \(n!\) is at least as large as the corresponding factor \(2\). Concretely, for \(n \ge 4\),

as one checks at \(n = 4\) (\(24 \ge 16\)) and which only improves as \(n\) grows. Taking reciprocals reverses the inequality:

The geometric series \(\sum \tfrac{1}{2^n}\) has \(r = \tfrac12\), so \(|r| < 1\) and it converges. Since the Comparison Test only requires the inequality to hold for all sufficiently large \(n\), it applies here and \(\sum \tfrac{1}{n!}\) converges.

Test the harmonic series \(\displaystyle\sum_{n=1}^\infty \frac{1}{n}\).

Solution.

Let \(f(x) = 1/x\). On \(x \ge 1\) this function is positive, continuous, and decreasing, so the Integral Test applies. Evaluating the improper integral as a limit:

The integral is infinite, so by the Integral Test the harmonic series diverges.

This result is worth remembering. The terms \(1/n\) tend to \(0\), yet the series still diverges — proof that \(\lim a_n = 0\) alone is never enough to conclude convergence.

Test \(\displaystyle\sum_{n=1}^\infty \frac{n^2}{2^n}\), a sum of the kind that arises when computing average energies from a partition function.

Solution.

Here \(a_n = n^2/2^n\). Form the ratio, remembering that dividing by \(a_n\) means multiplying by its reciprocal:

Since \(\frac{n+1}{n} = 1 + \frac1n \to 1\), we get

so the series converges by the Ratio Test. Notice the general lesson: an exponential in the denominator beats any power of \(n\) in the numerator.

Test

for convergence.

Solution.

Keep only the highest power of \(n\) in each part. In the numerator \(\sqrt{2n^2 - 5n + 1} \approx \sqrt{2n^2} = \sqrt{2}\,n\), and in the denominator \(4n^3 - 7n^2 + 2 \approx 4n^3\). So for large \(n\)

This suggests comparing with \(b_n = \dfrac{1}{n^2}\). Forming the required limit,

which satisfies \(0 < L < \infty\). Since \(\sum \tfrac{1}{n^2}\) is a \(p\)-series with \(p = 2 > 1\) it converges, so the given series converges by the Limit Comparison Test.

Test \(\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}\) for convergence.

Solution.

The terms alternate in sign, so no comparison or integral test applies directly. Take absolute values:

which is a \(p\)-series with \(p = 2 > 1\) and therefore converges by \(\eqref{eq:p-series}\). Since \(\sum|a_n|\) converges, the theorem above tells us the original series converges — and it converges absolutely.

Consider \(\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\) from \(\eqref{eq:alt-harmonic}\).

Solution.

Set \(a_n = 1/n\). Then \(a_{n+1} = \frac{1}{n+1} \le \frac1n = a_n\) for all \(n\), and \(\lim_{n\to\infty} 1/n = 0\), so both conditions in \(\eqref{eq:leibniz}\) hold and the series converges by the Alternating Series Test.

However, taking absolute values gives \(\sum 1/n\), the harmonic series, which diverges. The series therefore converges but not absolutely: it is conditionally convergent. Its sum is \(\ln 2\).

Find the radius and interval of convergence of

Solution.

The series is centred at \(a=-2\) with \(c_n = \dfrac{1}{\sqrt{n+1}}\). Apply the ratio formula for \(R\):

so \(\rho = 1\) and \(R = 1/\rho = 1\). The series therefore converges for \(|x+2| < 1\), that is \(-3 < x < -1\).

The endpoints must be checked one at a time, by substituting them into the original series:

• At \(x=-3\): \((x+2)^n = (-1)^n\), giving \(\sum \dfrac{(-1)^n}{\sqrt{n+1}}\). This alternates, its terms decrease, and \(1/\sqrt{n+1} \to 0\), so it converges by the Alternating Series Test. It does not converge absolutely (see the next endpoint), so convergence here is conditional.

• At \(x=-1\): \((x+2)^n = 1\), giving \(\sum \dfrac{1}{\sqrt{n+1}}\). This behaves like a \(p\)-series with \(p = \tfrac12 \le 1\), so it diverges.

One endpoint is included and the other is not:

Find the radius and interval of convergence of

Solution.

Here \(c_n = \dfrac{1}{n\,3^n}\), so

so \(\rho = \tfrac13\) and \(R = 1/\rho = 3\). Convergence therefore holds for \(|x-2| < 3\), that is \(-1 < x < 5\).

At the endpoints, substitute into the original series and let the \(3^n\) cancel:

• At \(x=-1\): \((x-2)^n = (-3)^n\), so the term is \(\dfrac{(-3)^n}{n\,3^n} = \dfrac{(-1)^n}{n}\). This is the alternating harmonic series, which converges — but only conditionally, since taking absolute values gives the divergent harmonic series.

• At \(x=5\): \((x-2)^n = 3^n\), so the term is \(\dfrac{3^n}{n\,3^n} = \dfrac{1}{n}\). This is the harmonic series, which diverges.

Hence the interval of convergence is

Find the Maclaurin series for \(\sin x\).

Solution.

Differentiate repeatedly. The derivatives of \(\sin x\) repeat with period four, so we only ever need these four:

and then the pattern \(0, 1, 0, -1\) repeats. Substituting into \(f(x)=\sum f^{(n)}(0)x^n/n!\), every even power drops out because its coefficient is zero:

Only odd powers survive, with alternating signs, so

The Ratio Test confirms \(R = \infty\): the factorial in the denominator grows faster than any power of \(x\), so the series converges for every \(x\).

A pendulum of length \(L\) swinging through an angle \(\theta\) from the vertical obeys

where \(g\) is the acceleration due to gravity. This equation cannot be solved in terms of elementary functions. Show how a series expansion rescues the situation for small swings.

Solution.

The obstacle is the \(\sin\theta\) on the right, which makes \(\eqref{eq:pendulum-exact}\) nonlinear. Replace it by its Maclaurin series:

For a small swing, \(\theta\) is small and each successive term is far smaller than the one before. Keeping only the first term gives the small-angle approximation \(\sin\theta \approx \theta\), and \(\eqref{eq:pendulum-exact}\) becomes

which is linear and has the familiar oscillating solution \(\theta(t) = \theta_0\cos(\omega t)\) with \(\omega = \sqrt{g/L}\). This is simple harmonic motion, and it is where the standard pendulum period \(T = 2\pi\sqrt{L/g}\) comes from.

How good is the approximation? The first term we discarded is \(\theta^3/6\), so the relative error is about \(\theta^2/6\). For a swing of \(10^\circ = 0.175\) radians that is roughly \(0.5\%\). Note that \(\theta\) must be in radians — the series is only valid in radians, which is the real reason radians are the natural unit for angles in physics.

Start with the geometric series for \(|x|<1\):

Differentiate term-by-term:

Integrate term-by-term:

Both series have radius \(R=1\).

To find the series for \((x+1)\sin x\), we multiply the Maclaurin series for \(\sin x\) by \((x+1)\):

Expanding and combining like terms:

Thus, the series begins

This illustrates how multiplying a series by a polynomial simply shifts and rescales its terms. Notice that this method is easier than computing successive derivatives of \((x+1)\sin x\) to form its Taylor series directly.

To find the series for \(e^x \cos x\), multiply the Maclaurin series for \(e^x\) and \(\cos x\):

Work power by power. For each power of \(x\), collect every way of forming it as (term from \(e^x\)) \(\times\) (term from \(\cos x\)):

Thus the series begins

Note that the \(x^2\) term cancels exactly. Two warnings: include every product contributing to a given power (the \(x^4\) row has three), and do not carry higher-order terms you will only discard — to reach \(x^4\) you never need a term beyond \(x^4\) from either factor.

Find the Maclaurin series for \(\tan x\) up to \(x^5\).

Solution.

Rather than differentiate \(\tan x\) five times, divide the known series

Since \(\tan x\) is odd and \(\tan 0 = 0\), only odd powers can appear. Write

and clear the denominator using \(\sin x = \tan x \cdot \cos x\):

Now match coefficients power by power:

Therefore

Unlike \(\sin x\) and \(\cos x\), this series does not converge for all \(x\): \(\cos x\) vanishes at \(x = \pm\pi/2\), where \(\tan x\) blows up, so \(R = \pi/2\). Dividing by a series always risks a smaller radius of convergence than either series started with.

Find the series for \(\dfrac{1}{1+x}\).

Solution.

Here \(p=-1\), so the coefficients are

Substituting into the binomial series — note that the signs are already carried by the coefficients, so we do not insert extra alternating signs by hand:

This is a useful check on the whole binomial machinery: the answer is just the geometric series with ratio \(-x\), which we already knew from Section 1.4.

Expand \(\sqrt{1+x} = (1+x)^{1/2}\).

Solution.

Here \(p=\tfrac12\), so \(\sqrt{1+x} = \sum_{n=0}^\infty \binom{1/2}{n} x^n\). Compute the first few coefficients directly from the definition, each being a product of \(n\) factors divided by \(n!\):

Hence

Only the first two terms are needed in most applications: for small \(x\), \(\sqrt{1+x} \approx 1 + \tfrac12 x\).

The exact potential energy of a mass \(m\) at height \(h\) above the surface of the Earth (radius \(R\), mass \(M\)) is

where \(G\) is the gravitational constant. The familiar classroom formula is \(U = mgh\). Reconcile the two.

Solution.

The exact expression depends on \(h\) in an awkward way. Factor out \(R\) so that the \(h\)-dependence sits in a small quantity:

Near the Earth's surface \(h \ll R\) — even a tall mountain has \(h/R \approx 0.001\) — so \(x = h/R\) is small and the binomial series with \(p = -1\) applies:

Therefore

The first term does not depend on \(h\); since only changes in potential energy are physically meaningful, it can be discarded by choosing where to put the zero of energy. The second term is exactly \(mgh\), and it identifies \(g = GM/R^2 \approx 9.8\ \text{m/s}^2\) — the surface gravity is not an independent constant of nature but the first series coefficient of the true \(1/r\) potential.

The next term is smaller by a factor \(h/R\), so for \(h = 1\) km the error in using \(mgh\) is about \(0.02\%\). This is the same pattern as the relativistic kinetic energy of Section 1.11: the simple formula is the first term of a series, accurate whenever the neglected ratio is small.

From \(\frac{1}{1-t} = \sum_{n=0}^\infty t^n\), substitute \(t = x^2\):

Find the Maclaurin series of \(e^{\tan x}\) up to \(x^4\).

First, use

Let \(u=\tan x\). Then

Compute powers of \(u\) to the needed order:

Substitute:

(The radius of convergence is limited by the nearest pole of \(\tan x\), so \(R=\pi/2\).)

Find the Maclaurin series for \(\arctan x\).

We use the identity

First, expand the denominator as a binomial series:

Now integrate term by term:

Thus,

The expansion of \(1/(1+t^2)\) is valid only for \(|t|<1\), so integration gives \(|x|<1\); the endpoints \(x = \pm 1\) survive because there the series becomes alternating with terms decreasing to zero. Setting \(x=1\) gives the celebrated

a beautiful formula but a hopeless way to compute \(\pi\): it converges so slowly that several hundred terms are needed for two decimal places.

Expand \(\ln x\) about \(x=1\):

Let \(h = x-1\), so \(x = 1+h\):

Thus:

Approximate \(\sqrt{1.1}\) using the binomial series.

Solution.

Write \(\sqrt{1.1} = \sqrt{1+x}\) with \(x = 0.1\), and use the expansion found above (the binomial series with \(p = \tfrac12\)):

Substituting \(x = 0.1\) term by term:

The true value is \(1.0488088\ldots\), so three terms already give an error of about \(4\times10^{-6}\). Observe how fast the terms shrink: each is roughly \(x/2\) times the one before. This is typical — when the expansion variable is small, a couple of terms suffice.

The exact relativistic kinetic energy of a particle of mass \(m\) moving at speed \(v\) is

Show that at everyday speeds this reduces to the familiar \(\tfrac12 m v^2\).

Solution.

Write \(x = v^2/c^2\), so that \(K = mc^2\left[(1-x)^{-1/2} - 1\right]\). For small \(x\),

Substituting and cancelling the \(1\):

The leading term is exactly the Newtonian kinetic energy. The first correction is smaller than it by a factor \(\tfrac34 v^2/c^2\), which for a jet aircraft (\(v \approx 300\) m/s) is about \(10^{-12}\) — utterly negligible. Newtonian mechanics is not a different theory here; it is the first term of a series.

A hot object emits light across a range of frequencies. Planck's law gives the energy radiated at frequency \(\nu\) by a body at absolute temperature \(T\) as

where \(h\) is Planck's constant and \(k\) is Boltzmann's constant. Before Planck, the accepted classical result was the Rayleigh–Jeans law \(u(\nu) \propto \nu^2 kT\). Show that Planck's law reduces to it at low frequencies.

Solution.

The physically meaningful comparison is between the photon energy \(h\nu\) and the thermal energy \(kT\). “Low frequency” means \(h\nu \ll kT\), so set

and expand the exponential in the denominator using the Maclaurin series for \(e^x\):

Note that the leading \(1\) cancels exactly; this is why the expansion is needed rather than simply setting \(e^x \approx 1\), which would give a useless \(0\) in the denominator. Substituting \(e^{h\nu/kT} - 1 \approx h\nu/kT\) into \(\eqref{eq:planck}\):

which is exactly the Rayleigh–Jeans law. So the classical result was never wrong, only incomplete: it is the leading term of Planck's law when \(h\nu \ll kT\). At high frequencies the neglected terms dominate, the exponential suppresses \(u(\nu)\), and Planck's law avoids the “ultraviolet catastrophe” — the classical prediction that a hot body radiates infinite total energy.