Find the smallest value of N that will make the inequality hold for all n N.
-$$a _ { n } = \sqrt [ n ] { n + 1 }$$
A) $\mathrm { L } = \ln 2 , \mathrm {~N} = 394$
B) $L = 1 , N = 652$
C) $\mathrm { L } = 1 , \mathrm {~N} = 394$
D) diverges

Question

Find the smallest value of N that will make the inequality hold for all n > N.
-$$a _ { n } = \sqrt [ n ] { n + 1 }$$
A) $\mathrm { L } = \ln 2 , \mathrm {~N} = 394$
B) $L = 1 , N = 652$
C) $\mathrm { L } = 1 , \mathrm {~N} = 394$
D) diverges

Quizplus · Accepted Answer

We can start by using the limit comparison test with $b_n = n^{1/n}$, which we know converges to 1: $$\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n+1}}{n^{1/n}} = \lim_{n\rightarrow\infty} \left(1 + \frac{1}{n}\right)^{1/n} = e^0 = 1$$ Since the limit is finite and positive, we know that $\sqrt[n]{n+1}$ and $n^{1/n}$ have the same behavior for large enough $n$. Therefore, we can use the fact that $n^{1/n}$ approaches 1 from above to find the smallest value of $N$ that makes the inequality hold for all $n>N$. If we choose $N=1$, then $n^{1/n}>1$ for all $n>1$. We want to find the smallest $N$ such that $\sqrt[n]{n+1} < N$, or equivalently, $(n+1) < N^n$ for all $n>N$. Let's try $N=2$ and see what happens: $$(n+1) < 2^n$$ $n=1$ satisfies the inequality, and for $n>1$ we can prove it by induction: Assume $(k+1) < 2^k$ for some $k>1$. Then: $$(k+2) = (k+1)+1 < 2^k + 1 < 2^k + 2^k = 2^{k+1}$$ Therefore the inequality holds for all $n>1$ and the smallest $N$ is indeed 2. However, the given answer choices are not consistent with this result. We can check (using a tool like WolframAlpha), that the limit of the sequence is indeed 1, so the answer is not D (diverges). Choices A and C both have $L=1$, but $N=394$ and $652$ respectively. We can check that $N=2$ is indeed the smallest value that works by finding the smallest $n$ such that $\sqrt[n]{n+1}\geq 2$. This occurs when $n=4$ (since $2^4=16$ and $5^{1/4}\approx 1.5$). Therefore, the correct answer is B.

Find the Smallest Value of N That Will Make the Inequality