What is ((2^n-1))/((2^{n+1)+1)}<= 2 ?

The solution to ((2^n-1))/((2^{n+1)+1)}<= 2 is True for all n

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((2^n-1))/((2^{n+1)+1)}<= 2

\frac{( 2 ^{n} - 1 )}{( 2 ^{n + 1} + 1 )} \leq 2

True for all n

Interval Notation

(- \infty, \infty)

Solution steps

\frac{( 2 ^{n} - 1 )}{( 2 ^{n + 1} + 1 )} \leq 2

Apply exponent rules

\frac{2 ^{n} - 1}{2 ^{n} \cdot 2 + 1} \leq 2

Let

u = 2^{n}

\frac{u - 1}{u \cdot 2 + 1} \leq 2

\frac{u - 1}{u \cdot 2 + 1} \leq 2 : u \leq - 1 or u > - \frac{1}{2}

u \leq - 1 or u > - \frac{1}{2}

Substitute back

u = 2^{n}

2^{n} \leq - 1 or 2^{n} > - \frac{1}{2}

2^{n} \leq - 1 :

No Solution

2^{n} > - \frac{1}{2} :

True for all

n

Combine the intervals

No Solution or True for all n

Merge Overlapping Intervals

True for all n

4 \cdot 3^{x} < 3^{x} + 6

3^{x} > 4

3^{x} > 2

2^{n} < 32

(\frac{1}{5})^{x} < 5^{5 x + 2}

What is ((2^n-1))/((2^{n+1)+1)}<= 2 ?
The solution to ((2^n-1))/((2^{n+1)+1)}<= 2 is True for all n