Summation with non upper limit

Question

Hi, I woild like assistance with showing work on how to solve a summation question And determining if it is convergent or divergent Thanks in advance! 

 

Question: (summation symbol)-3(4)^(n-1) 

n=2 

Answer 1

Hi Khloe:

This requires converting from sigma (summation) notation to a series and then summing the series. Here's how to do it.

Since it's nit clear what the n =2 line means, I'm assuming the summation is from n =1 to some number, N.

S_N = ∑{n = 1 to N} 3 x 4^n-1

= 3 x 4⁰ + 3 x 4¹ + 3 x 4² + . . . 3 x 4^N-1

= 3( 1 + 4¹ + 4² + . . . + 4^N-1) . . . . . . . . . taking out 3 as a common factor

Inside the parentheses is a geometric series, with first term 1, common ration 4, and number of terms = N. The sum to N terms of a geometric series with first term, a, and common ratio, r, is S_N = a(r^N - 1) / (r - 1)

∴ S_N = 3 x (4^N - 1) / (4 - 1)

= 3 x (4^N - 1) / 3

= 4^N - 1

Since 4^N gets larger as N gets larger, so does 4^N - 1.

Therefore, S_N goes to infinity as N -> infinity, that is, S_N diverges.