Solve an inequality involving powers of one quarter

Key takeaway: This is a nice little estimate problem. The term $(1/4)^{n+1}$ shrinks fast, so it helps to isolate the exponential first and then compare sizes carefully.

We want to solve

$$\frac{1}{n+1}\left(\frac14\right)^{n+1}\le \frac{1}{1000}.$$

A direct way is to test small integers, since the left side decreases very quickly as $n$ grows.

1) Try nearby values

For $n=3$:

$$\frac{1}{4}\left(\frac14\right)^4=\frac{1}{4^5}=\frac{1}{1024}.$$

And

$$\frac{1}{1024}\le \frac{1}{1000}$$

is false because $\frac1{1024}<\frac1{1000}$ is actually true? Let's compare correctly: since $1024>1000$,

$$\frac1{1024}<\frac1{1000},$$

so $n=3$ already works.

Check $n=2$:

$$\frac13\left(\frac14\right)^3=\frac{1}{192}$$

which is much larger than $\frac{1}{1000}$, so $n=2$ does not work.

2) Conclude the threshold

Because the left side decreases with $n$, all integers

$$n\ge 3$$

satisfy the inequality.

3) If you want an algebraic bound

Multiply both sides by $1000(n+1)4^{n+1}$:

$$1000\le (n+1)4^{n+1}.$$

The left side grows much slower than the right side, so once the inequality holds at one value, it stays true for larger $n$.

So the solution is

$$\boxed{n\ge 3}.$$

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Pitfalls the pros know 👇 It is easy to get turned around when comparing fractions like $1/1024$ and $1/1000$. Bigger denominator means smaller value. Also, if $n$ is assumed to be a natural number, make that explicit before listing the answer; otherwise the domain could shift the threshold.

What if the problem changes? If the right-hand side changed to $1/5000$, the same idea would still work, but the first valid $n$ would likely be larger. You would test small values until the exponential decay beats the new bound, then use monotonicity to extend it upward.

Tags: inequality, exponential decay, monotonic sequence