Problem 50
Question
Find the sum of the convergent series by using a well-known function. Identify the function and explain how you obtained the sum. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{3^{2 n-1}(2 n-1)} $$
Step-by-Step Solution
Verified Answer
The sum of the series is \( \arctan(\frac{1}{3}) \)
1Step 1: Identify the well-known function
The given series looks similar to the Taylor series expansion of \( \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots \) .
2Step 2: Draw parallels with Taylor series
The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{3^{2 n-1}(2 n-1)} \), which can be rewritten as \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{(2 n-1)3^{2 n-1}} \). This is very similar to \( \arctan(x) \) except that every x in \( \arctan(x) \) is replaced by \( \frac{1}{3} \) in the given series.
3Step 3: Use known function to find sum
Thus, we see that the given series is merely the series expansion of \( \arctan(x) \) evaluated at \( x = \frac{1}{3} \). Therefore, the sum of the given series equals \( \arctan(\frac{1}{3}) \).
Other exercises in this chapter
Problem 50
Use the series representation of the function \(f\) to find \(\lim _{x \rightarrow 0} f(x)\) (if it exists). $$ f(x)=\frac{\sin x}{x} $$
View solution Problem 50
Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(2 n) !}{n^{5}} $$
View solution Problem 51
(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0.0 \overline{75} $$
View solution Problem 51
Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{\sin n}{n}\)
View solution