Problem 58 - Essential Calculus

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{n}{(n+1) 2^{n-1}} $$

Step-by-Step Solution

Answer

The original series is convergent.

Step 1: Recognize the given series

First, acknowledge the series given to us is: \(\sum_{n=1}^{\infty} \frac{n}{(n+1) 2^{n-1}}\)

Step 2: Choose a suitable comparison series

Now, choose a series to compare the given one with. For this problem, a good choice would be the convergent geometric series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\)

Step 3: Apply the Limit Comparison Test

Now that a comparison series has been chosen, apply the Limit Comparison Test which states that if you have two series \(\sum_{n=1}^{\infty} a_n\) and \(\sum_{n=1}^{\infty} b_n\) where a sequence \(\{b_n\}\) is positive and \(\lim_{n \to \infty} \frac{a_n}{b_n} = L\) (where L is finite and positive), both series will either converge or diverge. Compute the limit as n approaches infinity for the ratio of the given series to the comparison series: \( \lim_{n \to \infty} \frac{(n/(n+1) 2^{n-1})}{(1/2^n)} \).

Step 4: Simplify the limit

Simplify the limit equation to obtain, \( \lim_{n \to \infty} \frac{n}{n+1} = 1 \).

Step 5: State the final conclusion

Given that \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) is a convergent series and our limit is finite and positive, by the Limit Comparison Test, we can conclude that the original series \(\sum_{n=1}^{\infty} \frac{n}{(n+1) 2^{n-1}}\) is also convergent.