Is $\frac{3^n-(-1)^n}{n}\ge 4$?

Question

Is $$\frac{3^n-(-1)^n}{n}\ge 4$$ for natural $n$?

My Attempt: Yes. The proof is by induction. Let $f$ be given by $f(n)=\frac{3^n-(-1)^n}{n}$. We have $f(1)=\frac{3+1}{1}=4\ge 4$. Assume that $f(r)\ge 4$ for some $r\in\Bbb N$. Clearly $f$ is an increasing function on $\Bbb N$ (since the increase in the exponentiation of $3$, plus or minus one, is greater than that of $n$). Thus $f(r+1)\ge f(r)\ge 4$. $\square$

I'm a bit iffy about the increasing part.

Answer 1

Exponential functions grow much much faster than polynomials. It is true that for sufficiently large real number $x>0$, $$ 3^x-1\geq 4x\tag{#} $$ In fact, one can show that (#) is true for all $x\geq 2$.⁽¹⁾ In particular, (#) is true for all positive integer $n\geq 2$. Note that (#) is stronger than your inequality.

Now, you only need to check your inequality when $n=1$.

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(1) To see why this is true, let $$ f(x)=3^x-1-4x. $$ Note that $f(2)=0$ and $f'(x)=3^x\cdot \log 3-4\geq 9-4>0$ for $x\geq 2$.

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proof-verification

Let $f$ be given by $f(n)=\frac{3^n-(-1)^n}{n}$. (This line is OK.)

We have $f(1)=\frac{3+1}{1}=4\ge 4$. (This line is OK.)

Assume that $f(r)\ge 4$ for some $r\in\Bbb N$. (This line is OK.)

Clearly $f$ is an increasing function on $\Bbb N$ (since the increase in the exponentiation of $3$, plus or minus one, is greater than that of $n$ (Oops, not clear, you need to demonstrate why.)).

Thus $f(r+1)\ge f(r)\ge 4$. $\square$

Answer 2

Note that

$$\frac{3^n-(-1)^n}{n}\ge 4 \iff3^n-(-1)^n\ge4n$$

thus since

$$3^n-(-1)^n\ge 3^n-1$$

we can prove by induction that for some $\bar n$ for $n\ge \bar n$ we have

$$3^n-1\ge4n$$

and then verify the cases $n<\bar n$ by direct calculation.

Notably for $3^n-1\ge4n$ we have

• base case $n=2\implies 3^2-1=8\ge8$
• inductive step let assume $3^n-1\ge4n$ then

$$3^{n+1}-1=3\cdot 3^n-1\ge3(4n+1)-1=12n+2\ge4(n+1)$$

Then $3^n-1\ge4n$ is true for every $n\ge 2$ and thus we need to verify that $3^n-(-1)^n\ge4n$ holds for $n=1$ which is true.

Answer 3

Since $3^n-(-1)^n>3^n-1$, it's enough to show that $$\frac{3^n-1}{n} \ge 4 \quad n \ge 2$$

(For $n=1$, you've already checked.)

So you need to prove $$f(n)=3^n-1-4n \ge 0 \; \forall n \ge 2 $$

We've $$f'(n) = 3^n \ln 3 -4$$

Now, we've that $f'(2)=9\ln 3-4 >0$ and $f'(n)$ is increasing too, since $3^n$ is increasing.

Therefore, $f(2)=0$ and $f'(n)>0 \; \forall n \ge 2$ proves that $$f(n) >f(2)=0 \implies \frac{3^n-1}{n} \ge 4$$