What is the sum from n=1 to infinity}(-1)^{n+1 of (3/(2^n)) ?

The sum from n=1 to infinity}(-1)^{n+1 of (3/(2^n)) is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity}(-1)^{n+1 of (3/(2^n))

n = 1 \sum \infty (- 1)^{n + 1} (\frac{3}{2 ^{n}})

converges

Solution steps

n = 1 \sum \infty (- 1)^{n + 1} \frac{3}{2 ^{n}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot n = 1 \sum \infty (- 1)^{n + 1} \frac{1}{2 ^{n}}

Apply Alternating Series Test:

converges

= 3 converges

= converges

\int \frac{1}{sin ( 2 y ) cos ( 2 y )} d y

\frac{d}{d x} (4 - e^{2 x})

\int (1 - cos (2 t)) (cos (t)) t d t

n = 0 \sum \infty n^{2} e^{- 7 n}

\int cos^{3} (X) sin^{2} (X) D X d X

What is the sum from n=1 to infinity}(-1)^{n+1 of (3/(2^n)) ?
The sum from n=1 to infinity}(-1)^{n+1 of (3/(2^n)) is converges