Answer to Question #205032 in Calculus for abcdefna

Question

Determine whether the following series converges or diverges. If the series converges, find its sum.

a. ∑∞𝑛=1 (1+2ⁿ⁺¹)/3n

b. ∑∞𝑛=1 (3+2ⁿ)/(2ⁿ⁺²)

c. ∑∞𝑛=1 1/(4n²-1)

Answer 1

a.

$\displaystyle\sum_{n=1}^{\infin}\dfrac{1+2^{n+1}}{3n}=\text{diverges}$

$\lim\limits_{n\to \infin}a_n=\lim\limits_{n\to \infin}\dfrac{1+2^{n+1}}{3n}=\infin\not=0$

The series diverges by the Test for divergence.

b.

$\displaystyle\sum_{n=1}^{\infin}\dfrac{3+2^{n}}{2^{n+2}}=\text{diverges}$

$\lim\limits_{n\to \infin}a_n=\lim\limits_{n\to \infin}\dfrac{3+2^{n}}{2^{n+2}}=\lim\limits_{n\to \infin}\dfrac{\dfrac{3}{2^{n}}+\dfrac{2^{n}}{2^{n}}}{\dfrac{2^{n+2}}{2^{n}}}$

$=\lim\limits_{n\to \infin}\dfrac{\dfrac{3}{2^{n}}+1}{4}=\dfrac{1}{4}\not=0$

The series diverges by the Test for divergence.

c.

$\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{4n^2-1}=\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{(2n-1)(2n+1)}$

$=\dfrac{1}{2}\bigg(\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{2n-1}-\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{2n+1}\bigg)$ $=\dfrac{1}{2}\bigg(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...\bigg)$

$\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{4n^2-1}=\dfrac{1}{2}$

The series converges and its sum is $\dfrac{1}{2}.$