Prove inverse $n$-th powers $< 1/a$ where $a=365$. Find $n$ for the series. Find $N$ so that for all $n\geq N$ holds that : 
$$
2^{-n} + 3^{-n} + 4^{-n} &lt \frac{1}{365}
$$
 A: A little playing around might help...
$$
2^{-n}+3^{-n}+4^{-n}&lt\frac1{365}
$$
$$
6^n+4^n+3^n&lt\frac{12^n}{365}
$$
$$
1+\left(\frac23\right)^n+\left(\frac12\right)^n&lt\frac{2^n}{365}
$$
The LHS decreases from $3$ to $1$ for $n\in\mathbb{N}$,
while the RHS increases from $\frac1{365}$ to $\infty$.
Graphing or trial and error soon finds the meeting point,
which over (by rootfinding) $\mathbb{R}$ is $8.55963662579$.
Therefore, $N=9$.
Even without rootfinding, we can answer this since $2^8=256$
so that the inequality is clearly false for $n=8$, while $2^9=512$
so that for $n=9$, the RHS becomes
$$\frac{512}{365}=1+\frac{147}{365}>1+\frac{512}{19683}+\frac{1}{512},$$
and without resorting to such high powers, one can note that
$$\left(\frac23\right)^2=\frac49&lt\frac12$$
so that the LHS satisfies
$$
1+\left(\frac23\right)^n+\left(\frac12\right)^n &lt 
1+\left(\frac12\right)^{n/2}+\left(\frac12\right)^n 
$$
and conclude
$$
1+\left(\frac23\right)^9+\left(\frac12\right)^9 &lt 
1+\left(\frac12\right)^4+\left(\frac12\right)^9 =
1+\frac1{16}+\frac1{512} &lt
1+\frac1{10} &lt \frac{512}{365}
\,.
$$
A: Note that if $n >0$ then $\dfrac{1}{3^n}&lt\dfrac{1}{2^n}$ and $\dfrac{1}{4^n}&lt\dfrac{1}{2^n}$. So 
$$\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}&lt\frac{3}{2^n}.$$
Thus if $\dfrac{3}{2^n}&lt\dfrac{1}{365}$, the desired inequality will hold. So let's make sure that $\dfrac{1}{2^n}&lt\dfrac{1}{3\cdot 365}$, or equivalently that $2^n>1095$. We are familiar with powers of $2$: certainly $2^{11}=2048>1095$. Therefore we can take $N=11$. 
Remark: There is a smaller $N$ that will work. But the question did not ask for the cheapest $N$ that works.  Sometimes, in doing estimates to prove convergence, students expend excessive effort finding  the smallest $N$ such that $|a_n-a|&lt\epsilon$ for all $n>N$. 
