Partial Sums of Infinite Series

Key Questions

  • How do you find the 4-th partial sum of the infinite series #sum_(n=1)^oo(3/2)^n# ?

    Answer:

    #195/16#

    Explanation:

    When dealing with a sum, you have a sequence that generates the terms. In this case, you have the sequence

    #a_n = (3/2)^n#

    Which means that #n#-th term is generates by raising #3/2# to the #n#-th power.

    Moreover, the #n#-th partial sum means to sum the first #n# terms from the sequence.

    So, in your case, you're looking for #a_1+a_2+a_3+a_4#, which means

    #3/2 + (3/2)^2 + (3/2)^3 + (3/2)^4#

    You may compute each term, but there is a useful formula:

    #sum_{i=1}^n k^i= \frac{k^{n+1}-1}{k-1}#

    So, in your case

    #sum_{i=0}^4 (3/2)^i= \frac{(3/2)^{5}-1}{3/2-1} = 211/16#

    Except you are not including #a_0 = (3/2)^0 = 1# in your sum, so we must subtract it:

    #sum_{i=0}^4 (3/2)^i = sum_{i=1}^4 (3/2)^i - 1 = 211/16 - 1 = 195/16#

    KillerBunny · 1 · May 13 2018
  • What is a partial sum of an infinite series?

    A partial sum #S_n# of an infinite series #sum_{i=1}^{infty}a_i# is the sum of the first n terms, that is,
    #S_n=a_1+a_2+a_3+cdots+a_n=sum_{i=1}^na_i#.

    Wataru · · Aug 30 2014

Questions

Tests of Convergence / Divergence