Understanding Engineering Mathematics

Problem 14.12

Apply the comparison test to the series

(i) U= 1 Y

1

2!

Y

1

3!

Y···

(

=e^1 − 1

)

(ii) U=

1

2

Y

1

4

Y

1

6

Y···

(i) As standard series take the geometric series

V= 1 +

1

2

+

1

22

+···

Clearly,ur≤ 1 vrand we know thatV converges because its common ratio is less

than 1, and soUalso converges.

(ii) U=^12 +^14 +^16 +···

TakeV= 1 +^12 +^13 +^14 +···, the harmonic series

Thenur=^12 vrand sinceV diverges, so doesU.

Alternating series

In the particular case of analternating series,suchas:

1 −^12 +^13 −^14 +···

where the signs alternate, there is a very simple test for convergence. Thus, let

S=u 1 −u 2 +u 3 −u 4 +···

(allui≥0). Then if

lim

n→∞

un=0andun≤un− 1 for alln>N

(Na certain finite number) then the series converges.

Problem 14.13

Examine the convergence of the series

S= 1 −^12 Y^13 −^14 Y···

In this case it is clear that

(i) limn→∞un= 0 (ii) un<un− 1 for alln

So the required conditions are satisfied and therefore the series converges. Notice the
difference a sign or two makes – if all the signs are positive, then we have the harmonic
series, which we know diverges.