How do you determine if #a_n=1-1.1+1.11-1.111+1.1111-...# converge and find the sums when they exist?

Question

Christopher Allers · Accepted Answer

#1.01010101...# #a_n = 1 -1.1 + 1.11 - 1.111 + 1.1111 - ...# #a_n = 1 + (-1.1 + 1.11) +(- 1.111 + 1.1111) +( - ...# #a_n = 1 + (0.01) +(0.0001) +( - ...# we can say that it squence is geometric progression where #a_1 = 1#, common ratio, # r = 0.01/1 =0.0001/0.01 = 1/100# sum of infinity, #s_oo = a_1/(1-r)# #S_oo= 1/(1 - 1/100) = 1/(99/100) = 100/99 = 1.01010101...#

Charles Anctil · Answer

undefined This is a series with alternating signs and increasing magnitude of terms. To determine if the series converges, we can use the Leibniz criterion for alternating series, which states that if the terms of an alternating series decrease in absolute value and approach zero, then the series converges.

Here, the terms are $a_n = 1 - 1.1 + 1.11 - 1.111 + \ldots$. Notice that each term is the sum of the previous term and a number between 0 and 1. This implies that the terms are approaching 1 (the limit of the series) but never actually reach it. However, the terms do decrease in absolute value as more terms are added.

Therefore, the series converges by the Leibniz criterion. To find the sum, we can consider the partial sums. Let $S_n$ denote the $n$th partial sum:

$S_1 = 1$,
$S_2 = 1 - 1.1 = -0.1$,
$S_3 = 1 - 1.1 + 1.11 = 0.01$,
$S_4 = 1 - 1.1 + 1.11 - 1.111 = -0.001$, and so on.

From these partial sums, we can see that $S_n$ approaches 0 as $n$ approaches infinity. Therefore, the sum of the series is 0.