Question

Determine the convergence or divergence of the series using any appropriate test from this chapter. Identify the test used. $$ \sum_{n=1}^{\infty} \frac{10 n+3}{n 2^{n}} $$

Short answer

According to ratio test, the series \( \sum_{n=1}^{\infty} \frac{10 n+3}{n 2^{n}} \) converges.

Step-by-step solution

Identify the Series Terms

Identify the terms of the series. In this case, the general term of the given series is \( a_n = \frac{10n+3}{n2^n} \). The next term \( a_{n+1} = \frac{10(n+1)+3}{(n+1)2^{n+1}} \)

Apply Ratio Test

Apply the ratio test, which is calculating the limit of the absolute ratio of consecutive terms as \( n \) tends to infinity. \( \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty} \left|\frac{\frac{10(n+1)+3}{(n+1)2^{n+1}}}{\frac{10n+3}{n2^n}}\right| \)

Simplify the Ratio

Simplify the ratio. \( \lim_{n \to \infty} \left|\frac{10(n+1)+3}{10n+3} \right| \times \frac{1}{2} \)

Compute the limit

Now we compute the limit \( \lim_{n \to \infty} \frac{10n+10+3}{10n+3} \times \frac{1}{2} = \frac{10}{2} = 0.5 \)

Draw conclusion

As the limit \( 0.5 < 1 \), according to the ratio test, the series converges.