TOPIC 6

Infinite series 1: Geometric and telescoping series

Main ideas.

Convergence and divergence: general definitions and intuitions • k Geometric series: k1=0 r • 1 Telescoping series k1= quadratic • P ⇤ P Exercises.

Exercise 6.1. For each of the series below, please Write out the first few partial sums S ,S ,S • 1 2 3 Write out a general formula for S • n Determine if the series converges. If the series is convergent, to what does it • converge? In either case, explain your reasoning.

1 2k 1 1 (1) ( 1)k (5) 3k k2 +2k Xk=0 Xk=1 1 5k 1 1 (2) (6) 4k k2 4 Xk=0 Xk=3 1 2 1 k (3) (7) 4k k2 + k Xk=0 Xk=1 1 1 5 (4) + 3k 6k Xk=0 ✓ ◆

Solution: (1) We have S0 =1 2 S =1 1 3 2 4 S =1 + 2 3 9 2 4 8 S =1 + 3 3 9 27

35 36 6. INFINITE SERIES 1: GEOMETRIC AND TELESCOPING SERIES

The general formula for Sn is

n k n k 2 n+1 n+1 k 2 2 1 3 3 2 Sn = ( 1) = = = 1 3k 3 1 2 5 3 k=0 k=0 3 " # X X ✓ ◆ ✓ ◆ n+1 As n large we have 2 0 and thus the series is convergent, ! 3 ! with 1 2k 3 ( 1)k = 3k 5 Xk=0 (2) We compute S0 =1 5 S =1+ 1 4 5 25 S =1+ + 2 4 16 In general we have

n k 5 n+1 n+1 5 1 4 5 Sn = = = 4 1 4k 1 5 4 k=0 4 " # X ✓ ◆ 5 n+1 Since 4 gets large as n gets large, the series does not converge. (3) We compute S0 =2 2 S =2+ 1 4 2 2 S =2+ + 2 4 16 The general formula for the partial sums is

n n k 1 n+1 n+1 2 1 1 4 8 1 Sn = =2 =2 = 1 . 4k 4 1 1 3 4 k=0 k=0 4 " # X X ✓ ◆ ✓ ◆ 1 n+1 Since 4 gets small as n gets large we see that the series converges and 1 2 8 = . 4k 3 Xk=0 (5) We first do a partial fraction decomposition, writing 1 1 1 1 1 = = . k2 +2k k(k + 2) 2 k k +2  Thus the partial sum Sn is given by 1 1 1 1 1 S = + + + + n 2 1 2 3 ··· n  1 1 1 . 3 4 ···n +3 6. INFINITE SERIES 1: GEOMETRIC AND TELESCOPING SERIES 37

Many terms, and we see that 1 1 1 1 S = 1+ . n 2 2 n +1 n +2  As n we have S 1 1+ 1 ; thus the series converges as !1 n ! 2 2 1⇥ 1⇤ 3 = . k2 +2k 4 Xk=1 (6) A partial fraction decomposition gives us 1 1 1 1 1 = = . k2 4 (k 2)(k + 2) 4 k 2 k +2  Thus the partial sums are 1 1 1 1 1 S = + + + + n 4 1 2 3 ··· n 2  1 1 1 . 5 6 ···n +2 There are 8 terms which do not cancel and thus 1 1 1 1 1 1 1 1 S = 1+ + + . n 4 2 3 4 n 1 n n +1 n +2  We can see that the sequence of partial sums converges and hence

1 1 1 1 1 1 = 1+ + + . k2 4 4 2 3 4 Xk=3  1 k (7) Clearly a factor of k cancels and the series is equivalent to k2 + k Xk=1 1 1 . k +1 Xk=1 We look at the partial sums and use the same grouping as in class: 1 1 1 1 1 1 1 1 S = + + + + + + + + . n 2 3 4 5 6 7 8 ··· n +1 1 1 > 2 > 2 | {z } | {z } 1 In particular, we notice that S2k 1 ends with + k and therefore ··· 2 1 1 1 1 S2k 1 = + + + > k. 2 2 ··· 2 2 k times 1 Since 2 k grows without bound,| we{z see that} the sequence of partial sums grows without bound and the series diverges.

Note: There are more problems on the next page! 38 6. INFINITE SERIES 1: GEOMETRIC AND TELESCOPING SERIES

Exercise 6.2. Write each of the following series in terms “standard” geometric series. Then determine whether it converges or not.

1 22k+1 Example: Suppose we are given the series . First we re-write as 3k Xk=2 2k+1 k k k 2 2 k 2 1 2 1 2 4 1 4 1 4 4 32 1 4 = · =2 =2 = . 3k 3k 3 3 3 9 3 Xk=2 Xk=2 Xk=2 ✓ ◆ Xk=2 ✓ ◆ ✓ ◆ Xk=2 ✓ ◆ Next we define a new counter l = k 2. Notice that when k = 2 our new counter has l = 0. Thus we have l 1 22k+1 32 1 4 = . 3k 9 3 Xk=2 Xl=0 ✓ ◆ 4 The series does not converge because the ratio 3 is greater than 1.

1 3n 1 3n (1) (4) ( 1)n 4n 2 4n n=2 n=3 X X · 1 2 1 1 2 (2) ( 1)n (5) 3n 3n 1 9n n=3 n=0 X X ✓ ◆ 1 2 1 5n 3 (3) (6) 3n+2 62n 1 n=1 n= 1 X X

Solution: (1) We write the series as n 1 3n 1 3 = . 4n 4 n=2 n=2 X X ✓ ◆ 3 Already we see that the series converges, as the geometric ratio 4 is less than 1. We now re-index, setting k = n 2 and obtaining k+2 2 k 1 3n 1 3 3 1 3 = = . 4n 4 4 4 n=2 X Xk=0 ✓ ◆ ✓ ◆ Xk=0 ✓ ◆ Optional: At this point we can easily determine the precise value that the series converges to; we find 2 1 3n 3 1 = 4n 4 1 3 n=2 4 X ✓ ◆ 6. INFINITE SERIES 1: GEOMETRIC AND TELESCOPING SERIES 39

1 2 (2) ( 1)n We may write the series as 3n n=3 X n 1 2 1 1 ( 1)n =2 . 3n 3 n=3 n=3 X X ✓ ◆ 1 Since 3 < 1 the series converges. Re-indexing, we set l = n 2 and obtain 2 l 1 2 1 1 1 ( 1)n =2 . 3n 3 3 n=3 X ✓ ◆ Xl=0 ✓ ◆ Optional: At this point we can easily determine the precise value that the series converges to; we find 2 1 2 1 1 2 3 1 ( 1)n =2 = = . 3n 3 1+ 1 3 · 4 2 n=3 3 X ✓ ◆ 1 5n 3 (3) We write the sum as 62n 1 n= 1 X n 1 5n 3 5 3 1 5 = 62n 1 6 1 36 n= 1 n= 1 ✓ ◆ X X from which we see that the series converges. Re-indexing with k = n +1wefind n 3 3 k 1 3 k 1 5 5 1 5 6 1 5 = = . 62n 1 6 1 36 54 36 n= 1 k=0 ✓ ◆ k=0 ✓ ◆ X X X Optional: Using the re-indexed expression for the sum we find

1 5n 3 63 1 = 62n 1 54 1 5 n= 1 36 X Exercise 6.3. (Challenge) Recall that the geometric sum formula gives us a nice expression for the quantity

1+x + x2 + + xn. ··· Use this to find a nice formula for the quantity

n n 1 1+2x +3x + + nx . ··· Use your formula to analyze the convergence of the series n 1 1 1 k . 3 Xk=1 ✓ ◆