In considering whether the sequence defined by 1 / [ `sqrt(n) * ln(n) ] converges or diverges, it was suggested that since 1 / `sqrt(n) and 1 / ln(n) both diverge, then their product 1 / [ `sqrt(n) * ln(n) ] should diverge.

This reasoning does not stand up to scrutiny. Consider the following example:

• 1 / n^.75 diverges, being a p series with p < 1.
• 1 / `sqrt(n) also diverges for the same reason.
• By the above reasoning, the product series 1 / n^.75 * 1 / `sqrt(n) should also diverge.
• However, the product series is just the same as 1 / n^1.25, which being a p series with p > 1 converges.

To determine whether the series 1 / [ `sqrt(n) ln(n) ] converges or diverges, we might suspect that the series diverges and compare it with 1 /n.

• We ask whether, beyond some value of n, n ln(n) < n; if so, then the denominator of our series is less than the denominator of the 1 / n series and the terms of our series will be greater than those of the divergent 1 / n series. We will therefore conclude that our series is diverge.
• Our inequality will certainly hold for sufficiently large values of n provided the indicated limit is infinite.

To prove that the limit as infinite, we used l 'hopital's rule.

• The ratio of the derivatives is `sqrt(x) / (1 + ln(x) ), which we easily see to be infinite.

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Thus our original inequality does hold, the terms of our sequence eventually are always greater than those of the 1/ n sequence, and our sequence diverges.

Alternatively, we can show that the series 1 / [ n ln(n) ] diverges using an integral comparison test.

• We can then use a comparison test to show that 1 / [ `sqrt(n) ln(n) ] also diverges.

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The series defined by 1 / [ n^2 ln(n) ] can be proved to be convergent by means of a comparison with the convergent p series 1 / n^2.

• Since for most values of n, ln(n) > 1, 1 / [ n^2 ln(n) ] < 1 / n^2 and the series converges.
• More specifically, for n > e, ln(n) > 1. Since e < 3, then for n > 3 we have ln(n) > 1. The comparison therefore holds for n > 3, in this proves that our series is convergent.

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As we have seen, when a linear restoring force F = - k x tends to move an object of mass m back to its equilibrium position, then in the absence of any other forces the object will postulate about its equilibrium position with angular frequency `omega = `sqrt(k / m), with the indicated position function x(t).

• The velocity v(t) will then be the derivative of the position function x(t), and acceleration will be the second derivative of the position function.
• The two constants A and B are often evaluated by the initial conditions on x and v.

The first figure below depicts a pendulum with net force F = - kx, and the differential equation from which the above results were obtained.

• In this case, we are assuming that there is no other force acting on the pendulum.
• Specifically we are therefore assuming that there is no air resistance or any other type of 'drag force' acting on the pendulum.
• A damping force is one that dissipates the energy of the pendulum's motion. In the present case we say that the pendulum is undamped.

In the real world a pendulum will typically encounter air resistance.

• This air resistance depends on the velocity of the pendulum.

Air resistance is a very complex phenomenon and can only be modeled approximately.

• Typical models, depending on the size and shape of the moving object and its speed, assume their resistance proportional to the first, second or third power of the velocity.

In this case we will assume a resistance proportional to the first power of the velocity, so that the resistance force is - c v. The negative sign indicates that the 'drag' force is in the opposite direction to the velocity.

• With this drag force the net force on the object is F = - kx - cv.
• Sense F = m a = m d^2 x / dt^2 and v = dx / dt, we obtain the differential equation in the second figure below.

To solve this equation we begin by placing all nonzero terms on the left-hand side of the equation.

• We might also divide both sides of the equation by m, so that the second derivative term has coefficient 1.

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Since we have a sum of constant multiples of the derivatives of the function x(t) equal to 0, our solution x(t) might well have the property that all its derivatives are multiples of x(t).

• We therefore try a solution of the form x = C e ^ (rt).
• The derivatives of this function are as indicated, and as expected all are constant multiples of the original function.

We substitute these derivatives into the originally equation and obtain the equation at the bottom of the figure below.

We divide both sides of the equation by C e ^ (rt) and obtain an equation for r.

• The quadratic equation is easily solved, as indicated.
• We obtain two solutions r1 and r2, resulting in two functions C1 e ^ (r1 t) and C2 e ^ (r2 t) that satisfy the differential equation.
• Is easily verified that the sum x(t) = C1 e ^ (r1 t) + C2 e ^ (r2 t) of these two functions also satisfies the differential equation.

We note that there are three possible types of solutions to the quadratic equation, depending on whether the discriminant c^2 - 4 k m is positive, negative, or zero.

• When there is a lot of drag and neither the mass nor the force constant is very great (more specifically, when the product of mass and force constant isn't too large), c^2 - 4 k m will be positive and we will obtain two real solutions r1 and r2, and our differential equation will have solution x(t) = C1 e^(r1 t) + C2 e^(r2 t). 

• We call this case the 'overdamped' case; the damping overcomes the force constant and mass and dominates the behavior of the system.

• When there is very little drag and/or when mass and force constant are large, c^2 - 4 k m will be negative and we will obtain two complex conjugate solutions.  We will see later how to deal with complex solutions.

• We call this case the 'underdamped' case; the damping has its effect, but the behavior is also much like that of an undamped oscillator.  We end up with a sine or cosine position function, but with exponentially decreasing amplitude.

• The third case, where the discriminant c^2 - 4 k m = 0, is called the 'critically damped' case.  It is on the borderline between the underdamped and overdamped cases.

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As an example we find a solution for c = 4 and k = 1, m = .5.

• We note that these constants correspond to a fairly large drag coefficient c and fairly small values of the restoring force constant k and the mass m. We could think of this as representing a pendulum with a relatively small mass in a fairly viscous fluid.

In this case the solutions of the quadratic formula yield r1 = - 7.7 and r2 = -.26.

• The solution for x(t) is therefore as indicated.

If we assume initial conditions x(0) = 2 and v(0) = 3, then we will obtain from these conditions the equations C1 + C2 = 2 and -7.7 C1 - .26 C2 = 3.

• The second equation requires us to first find the function v(t) = x ' (t) = -7.7 C1 e^(-7.7 t) - .26 C2 e^(-.26 t); substitution for v(0) then yields the second equation.
• Our solutions are C2 = 2.5, C1 = -.5, leading to the indicated solution x(t) = -.5 e^(-7.7 t) + 2.5 e^(-.26 t).

You should graph these two exponential functions on the same set of coordinate axes.

• You will find that the first function intercepts the vertical axes at -.5, the second at + 2.5.
• You will find also that the first function decays much more rapidly than the second.
• The first function therefore represents transient behavior of the solution, i.e., behavior which is significant only for small values of t, while the second represents the longer-term decay of the position function.