How to find the sum of this series?

Question

The following series satisfies the hypothesis of the alternating series test
#1/(19) - 1/(29^2)+1/(39^3)-1/(49^4)+...#
Approximate the sum of the series to two decimal place accuracy.

Luke Alvarez · Accepted Answer

#1/(1*9) - 1/(2*9^2) + 1/(3*9^3) - 1/(4*9^4) + ... = ln(10) - ln 9#.

We have #S = 1/(1*9) - 1/(2*9^2) + ... = sum_(z=1)^(oo) (-1)^(z-1)/(z*9^z)# Let's solve the general case instead, as that's going to be easier to understand. #epsilon = n/1 - n^2/2 + n^3/3 -...# If you plug #n = 1/9#, you'll get #S#. Notice how #epsilon# is a alternating sum with the general term #n^k/k#. Aha! This looks a lot like the antiderivative of #n^(k-1)#. Here, calculus comes into play. By plugging in #n^k/k = int n^(k-1)dn#, you'll get #epsilon = int n^0 dn - int n^1dn + int n^2 dn -...# #epsilon = int (n^0 - n^1 + n^2-...)dn# Let's separate the positive powers and negative powers, respectively : #epsilon = int(n^0 + n^2 + n^4 +...)dn - int(n^1 + n^3 + ... n^5+...)dn# #epsilon = int(n^0 + n^2 + n^4 +...)dn - int n(n^0 + n^2+n^4 +...)dn# Let #t=n^2#. #epsilon = int(1+t+t^2+...)dn - int n(1+t+t^2+...)dn# We are #"not"# substituting #t# into the integral. We just want the infinite sum to be simpler to look at. Now, we know that #1+t+t^2+... = "first term" * (1-"ratio"^"number of terms")/(1-"ratio")# The first term is #1#, the ratio is #t# and the number of terms is #eta#, but we have to take the limit as #eta -> oo#. #lim_(eta->oo) 1+t+t^2 + ... + t^eta = (1-t^eta)/(1-t)# This is important; the geometric series we have only converges if #| t| < 1#, and so does #epsilon#. If #|t| <1#, then #lim_(eta->oo) t^eta = 0#. #lim_(eta->oo) 1+t+t^2 + ... t^eta = 1/(1-t)# #1+t+t^2+... = 1/(1-t)#. Back to our original sum : #epsilon = int (color(red)(1+t+t^2+...))dn - int n(color(red)(1+t+t^2+...))dn# #epsilon = int color(red)(1/(1-t)) dn - int n/color(red)(1-t)dn# #epsilon = int (1-n)/(1-t)dn# Since #t = n^2#, we have #epsilon = int (1-n)/(1-n^2) dn = int (1-n)/((1-n)(1+n))dn# #epsilon = int 1/(1+n) dn# This is a fairly simple integral. Let #u = 1+n => du = dn#. #epsilon = int 1/u du# The antiderivative of #1/u# is equal to the natural logarithm of #u#. #int 1/u du = ln u# #=> epsilon = ln u = ln (n+1)# Not only did we get the infinite series for #ln (n+1)#, but also the answer we were looking for : #n/1 - n^2/2 + n^3/3 -... = ln(n+1)#. In our case, #n = color(red)(1/9#. #1/(1*color(red)9) - 1/(2*color(red)9^2) + 1/(3*color(red)9^3) -... =# #= ln(1/9 +1 ) = ln (10/9) = color(red)(ln 10 - ln 9)#. In order to approximate it, we have to use a formula : #ln(k) = log_10 k / log_10 e ~~ log_10 k/0.4343# From now on, instead of #log_10k# I'll be using just #logk#. #ln(10/9) ~~ log(10/9)/0.4343 = (log10-log9)/0.4343# #ln(10/9) ~~ (1-log 9)/0.4343# Note that #0.95 < log 9 < 0.96#. Plugging these values in, we get #ln(10/9) ~~ (1-0.95)/0.4343 ~~ 0.115 = l_1# #ln(10/9) ~~ (1-0.96)/0.4343 ~~ 0.092=l_2# The inequality above helps us deduce that, consequently , #ln(10/9) ~~ (l_1+l_2)/2 ~~ 0.103#. The actual answer is #0.105#, but we only need #color(red)("two")# decimal places. Finally, #color(red)(1/(1*9) - 1/(2*9^2) + 1/(3*9^3) - 1/(4*9^4) + ... ~~ 0.10 "."#

How to find the sum of this series?

The following series satisfies the hypothesis of the alternating series test #1/(1*9) - 1/(2*9^2)+1/(3*9^3)-1/(4*9^4)+...# Approximate the sum of the series to two decimal place accuracy.

The following series satisfies the hypothesis of the alternating series test
#1/(19) - 1/(29^2)+1/(39^3)-1/(49^4)+...#
Approximate the sum of the series to two decimal place accuracy.