Question

Find the sum of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{3^{n}(2 n+1)} $$

Short answer

The sum of the series is 1.

Step-by-step solution

Identify the first term and the common ratio

The first term a is obtained by letting n=0, which results in \((-1)^0/(3^0(2*0+1)=1\). Further, the common ratio r in this case is complicated by the inclusion of n in the denominator. Instead of being -1/3 as in a basic geometric series it will be \(-1/3 *(2n+1)/(2n+3)\) when we simplify the ratio of the generic (n+1)th term to the nth term.

Formulate a sequence for partial sums

A common technique for summing a complicated series is to find an expression for the nth partial sum (the sum of the first n terms) and then take the limit as n approaches infinity. Being an infinite alternating series, partial sums will help to track convergence or divergence. So, let's create a generic partial sum sequence S(n) = 1 - 1/3(3/5) + 1/9(5/7) - .... +/- 1/3^n[(2n+1)/(2n+3)]. Observe that each term (from the second onwards) cancels out a part of the previous term. This is very unique to this series and we should note it.

Calculate the sum of the series

Using the cancelation pattern in the partial sum sequence, we can rewrite the sequence as S(n) = 1 - [1/3(2/5) - 1/3^2(4/7) + ... +/- 1/3^n(2n/(2n+3))]. This can be simplified to S(n) = 1 - 1/3^n * (2n/(2n+3)). Taking the limit as n approaches infinity, we get the expression 1 - 0 that equals to 1 which represents the sum of the series.