Problem 55 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The sum of the series is 1.

Step 1: 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.

Step 2: 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.

Step 3: 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.