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Math 122 Lab 4: Calculator Infinite Series

How Good is the Calculator?

Our class topic has shifted to infinite series, so this lab focuses on the calculator’s abilities there. As you will see, this is an area where the calculator is not all that helpful. Determining convergence or divergence is still something that humans are better at than the TI-89!

What to Turn In

Turn in this worksheet with everything filled in. This will be graded for accuracy and understanding, so write good explanations when you are asked to explain something.

Geometric Series

This is the easiest type of series for us to handle and the easiest for the calculator. First, try the geometric series S_k=0^¥ 3(1/2)^k. This series converges because r=1/2 and |r|<1. We also know that it converges to c/(1-r) = 3/(1-1/2) = 6. On the TI-89, the Sigma key is found in the Calculus menu (F3) command number 4. The infinity key is the green option on the Catalog key. The full command should look like (using the letter z; you can use any letter you like)

S ( 3(1/2)^z,z,0,¥ )

Write the answer here. Is it correct?

Now try S_k=0^¥ 3(-1/2)^k. Write the answer here. Is it correct?

Next try S_k=0^¥ 3(2)^k. Write the answer here. Is it correct? Briefly explain.

Finally, try S_k=0^¥ 3(-2)^k. Write the answer here. What happened this time?

All in all, how does the calculator do on geometric series?

Harmonic Series

The next fact we learned is that the harmonic series diverges. Let’s see if the calculator knows this one. Try S_k=1^¥ 1/k. Write the answer here.

Integral Test

We have also looked at the integral test. Using this test, we can show that S_k=1^¥ 2k/(k² +1) diverges. Try this series on the TI-89 and write the answer here.

Well, the calculator can do integrals. Try ò₁^¥2x / (x² +1) dx and write the answer here. What does this answer tell you about the convergence or divergence of the series?

Recall that the integral test requires that the function (in this case, 2x / (x² +1) ) must be positive and decreasing. It is clearly positive, but is it decreasing? Recall that a decreasing function has a negative derivative. Compute the derivative on the TI-89, write the answer here and explain why you know that it’s negative for x>1. (This is all that’s required since the series starts at k = 1).

p-Series

As we have seen, the integral test lets us know the convergence or divergence for p-series. They have the form S_k=1^¥ 1/k^p. We’ll try some p-series on the calculator. First try S_k=1^¥ 1/k² and write the answer here.

Next try S_k=1^¥ 1/k³ and S_k=1^¥ 1/k⁴ and write the answers here. What do you think is happening with p = 3?

Next try S_k=1^¥ 1/k⁵ and S_k=1^¥ 1/k⁶ and write the answers here. What do you expect will happen if we try other odd values p?

Next try S_k=1^¥ 1/k⁸ and S_k=1^¥ 1/k¹⁰ and write the answers here. Discuss the pattern of series values for even p.

Riemann zeta function

You can think of the series S_k=1^¥ 1/k^p as a function of p. This function is called the Riemann zeta function and is the subject of the most famous unsolved problem in modern mathematics. Three popular books have been published on attempts to solve this problem. One is called Prime Obsession by author John Derbyshire. You’ll see in exercise 59 on page 612 what this infinite series function has to do with prime numbers. (There’s nothing for you to do here, but you might want to go buy the book or look it up on the web.)

Approximating p-Series

We’ve seen that the calculator can’t compute the sum of a p-series for odd p’s. However, it can compute partial sums fairly rapidly. The syntax is the same as before, but we’ll replace the ¥ with different values of n. Try

S ( 1/z^3.,z,1,100 )

and write down the answer. Change the 100 to 200 and then 300. Keep changing to larger n’s (write down all answers) until you’re confident that you know the answer correct to 3 decimal places. If the sum equals π³ / n for some integer n, what is your best estimate of n?

Keep in mind that the calculator will compute partial sums for any series, whether it converges or diverges. Try partial sums of the harmonic series with n = 100, then 200, then 300. Write down the answers here. Is there anything about these calculations that might lead you to believe that this is a divergent series?

Repeat the above process for p-series with p = 0.9 and p = 1.1. You know which one converges and which one diverges, but if you didn’t know, what difference is there in the partial sums? Be as honest as possible. How confident should you be using partial sums to determine convergence or divergence?

There is an important approximation to partial sums of the harmonic series. The sum 1+1/2+1/3+…+1/n is approximately equal to ln(n)+ g where the constant g (the Greek letter gamma) is called Euler’s constant. Its definition is

g = lim_n®¥ ( 1+1/2+1/3+…+1/n – ln(n) )

Compute approximations of g for large values of n until you have three decimal places correct.

Exercises #25, 28, 29-32, 37-42 on page 682 can be worked using techniques from sections 7.2 and 7.3. Quickly identify which technique(s) you can use (geometric, harmonic, Divergence Test, Integral Test, Comparison Test, Limit Comparison Test, p-series) and then use one technique to conclude whether the series diverges or converges.