What is the limit as n approaches infinity of |((2n)^{2n})/((2n+2)^{2n)}| ?

The limit as n approaches infinity of |((2n)^{2n})/((2n+2)^{2n)}| is 1/(e^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as n approaches infinity of |((2n)^{2n})/((2n+2)^{2n)}|

n \to \infty lim (\frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}})

\frac{1}{e ^{2}}

Decimal Notation

0.13533 \dots

Solution steps

n \to \infty lim (\frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}})

\frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}}

is positive when

n \to \infty

. Therefore

\frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}} = \frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}}

= n \to \infty lim (\frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}})

Simplify

\frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}} : (\frac{2 n}{2 n + 2})^{2 n}

= n \to \infty lim ((\frac{2 n}{2 n + 2})^{2 n})

\frac{2 n}{2 n + 2} = \frac{n}{n + 1}

= n \to \infty lim ((\frac{n}{n + 1})^{2 n})

Apply exponent rules

= n \to \infty lim (e^{2 n l n (\frac{n}{n + 1})})

Apply the Limit Chain Rule:

\frac{1}{e ^{2}}

= \frac{1}{e ^{2}}

x \to 2 lim (3 x - 1)

\frac{d y}{d t} = 6 y + 2 t - 4

\int \frac{- 2}{2 x - 3} d x

n = 1 \sum \infty 5^{n} x^{n}

arctan (4 x)

What is the limit as n approaches infinity of |((2n)^{2n})/((2n+2)^{2n)}| ?
The limit as n approaches infinity of |((2n)^{2n})/((2n+2)^{2n)}| is 1/(e^2)