Series

Greg Warrington - 08.05.05

with(plots): with(plottools):

Alternating Harmonic Series

hssums := plot([seq([t,sum((-1)^(k-1)/k,k=1..t)],t=1..50)], style=point,thickness=3,color=black):

This animation shows how the partial sums of the alternating harmonic series converge to a specific

value that looks to be a little less than 0.7 (in fact, it turns out to be ln(2) = 0.693...).

Click on the picture and then move the slider to take partial sums with successively more terms.

animate( pointplot, [ [[t,sum((-1)^(k-1)/k,k=1..t)]], symbol=box, symbolsize=30,thickness=30],t=1..50, frames=50, background=hssums);

Harmonic Series

ahssums := plot([seq([t,sum(1/k,k=1..t)],t=1..50)],style=point,thickness=3,color=black):

This is similar to the previous animation, though it uses the usual Harmonic Series rather than the

Alternating version. Notice that the partial sums keep getting larger and larger without settling down.

animate( pointplot, [ [[t,sum(1/k,k=1..t)]], symbol=box, symbolsize=30,thickness=30],t=1..50, frames=50, background=ahssums);

esums := plot([seq([t,sum(1/k!,k=0..t)],t=2..10)],style=point,thickness=3,color=black):

This animation shows the partial sums

Sum(1/k!,k=0..t);

As t runs from 2 to 10. These partial sums settle down very quickly to something a little larger than 2.7.

(In fact, the partial sums settle down to e = 2.71828....)

animate( pointplot, [ [[t,sum(1/k!,k=0..t)]], symbol=box, symbolsize=30,thickness=30],t=2..10, frames=10, background=esums);

Computing sums with Maple

sum(1/k,k=1..infinity);

sum(1/k,k=1..20);

evalf(%);

sum((-1)^(k+1)/k,k=1..infinity);

sum(1/(1+k^2),k=0..infinity);

4*sum((-1)^(k+1)/(2*k-1),k=1..infinity);

Sometimes Maple can't compute a sum exactly even when the formula is known.

Consider the following horrific sum and its numeric value.

12*Sum(((-1)^k*(6*k)!*(13591409+545140134*k))/((k!)^3*(3*k)!*(640320^3)^(k+1/2)),k=0..infinity);

evalf[30](%);

evalf[30](1/Pi);

There are certain families of series that evaluate to numbers that are very closely related.

The below examples motivate why it might be useful to consider "polynomials of infinite degree"

Sum(1/k!,k=0..infinity);

Sum((-1)^k/k!,k=0..infinity) = sum((-1)^k/k!,k=0..infinity);

Sum(3^k/k!,k=0..infinity) = sum(3^k/k!,k=0..infinity);

Sum(x^k/k!,k=0..infinity);