Slowly Convergent Series

by
K. D.  Stroyan
University of Iowa

copyright 1997 by Academic Press, Inc. - All rights reserved.

Notebook Overview

This NoteBook shows that the next term in the series

1 + (1/4) + (1/9) + ⋯ + (1/(n^2)) + ⋯

is NOT a good estimate of the error in summing only  n  terms.  The computer doesn't prove what we need but may help you see how to make the needed estimates.

References to the Text

This NoteBook goes with Exercise 27.2.2 of the text.

Step-by-step help with Exercise 27.2.2

In the Chapter 26 NoteBook SomeSums we show that

((π^2)/6) = 
        1 + (1/4) + (1/9) + ⋯ + (1/(n^2)) + ⋯

Here we estimate the error in adding only to the  nth  term of the series.  The moral of the story is that the error is NOT even close to the size of the next term  (1/(((n + 1))^2))

1) Two telescoping sums

Prove that

(∑_(k = 1)^n) (1/(k (k + 1))) = 1 - (1/(n + 1))

1 + (∑_(k = 2)^n) (1/(k (k - 1))) = 1 + 1 - (1/n)

HINT: Verify that  (1/(k (k + 1))) = (1/k) - (1/(k + 1))  and  (1/(k (k - 1))) = (1/(k - 1)) - (1/k) and sum these instead.  ((1/2) - (1/3)) + ((1/3) - (1/4)) + ((1/4) - (1/5)) + 
        ⋯ ???
Here are more hints:

Clear[x, loSum, midSum, hiSum]; loSum[n_] := (∑_(k = 1)^n) (1/(k (k + 1))); midSum[n_] := (∑_(k = 1)^n) (1/(k^2)); hiSum[n_] := 1 + (∑_(k = 2)^n) (1/(k (k - 1))); loSum[n]midSum[n]hiSum[n]n = 9Table[(1/(k (k + 1))), {k, 1, n}]Table[{(1/k), (1/(k + 1))}, {k, 1, n}]loSum[n]1 - (1/(n + 1))Table[(1/(k^2)), {k, 1, n}]midSum[n]coefs = {1}; 
    Flatten[AppendTo[coefs, Table[(1/(k (k - 1))), {k, 2, n}]]]Table[{(1/(k - 1)), (1/k)}, {k, 1, n}]hiSum[n]1 + 1 - (1/n)

2) Lower and Upper Estimates

Prove that

(1/(k (k + 1))) ≤ (1/(k^2)) ≤ 
        (1/(k (k - 1))),  for  k ≥ 2

so that

1 - (1/(n + 1)) = loSum[n] < midSum[n] < hiSum[n] = 2 - (1/n)

3) Lower and Upper Sum Estimates

Show that

(∑_(k = n + 1)^∞) (1/(k (k + 1))) < 
        (∑_(k = n + 1)^∞) (1/(k^2)) < 
        (∑_(k = n + 1)^∞) (1/(k (k - 1)))

so

(1/(n + 1)) < (∑_(k = n + 1)^∞) (1/(k^2)) < (1/n),  for n > 2

4) What's the Difference?

Show that because of part (3), we can write

(∑_(k = n + 1)^∞) (1/(k^2)) = 
        (1/(n + 1)) + ϵ

with  0 < ϵ < (1/n) - (1/(n + 1)).

5) ϵ < (1/(n^2))

Prove that epsilon from part (4) is less than  (1/(n^2))  and that  (1/(n + 1)) + ϵ < (1/n)

7) Comparisons

Compare the actual limit of the sum of (1/(k^2))  to the sum of  100  terms.  Calculate the difference and compare that with  (1/n) + ϵ and with

N[((π^2)/6)]n = 100N[((π^2)/6) - midSum[n]]N[(1/n)]NSum[(1/(k^2)), {k, n + 1, 10 n}]N[ .(1/n)]N[(1/(((n + 1))^2))]

7)  Do the loSum and hiSum approximate midSum?

The ERROR in summing the loSum or hiSum compares with the ERROR in summing midSum only to  n  terms.  Show that the sums themselves, however, are NOT close to one another.