Study the sequence $x_n=\sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}$.

Question

I am given the sequence: $$ x_n=\sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}} $$ Where $n \ge 2$. I have to describe the sequence $(x_n)_{n \ge 2}$, where I am given the following options (I have to choose one):

A. convergent

B. bounded and divergent

C. unbounded and divergent

D. has negative terms

E. has infinite limit

What I tried was to use something like:

$$y_n = \ln x_n$$

$$y_n = \dfrac{\ln(2^{n \sin 1} + 2^{n \sin 2} + 2^{n \sin 3} + ... + 2^{n \sin n})}{n}$$

But I got stuck here and I can't find the limit. I should probably use a different approach, but I can't think of any. I can't continue what I've started either.

How should I approach this?

By the way, is choice B just something to trick me, or is it an actual possibility? Can a sequence be bounded and divergent? That doesn't seem right...

Answer 1

Let $$ a_n=\max\{\sin k: k=1,\ldots,n\}. $$ Then $\{a_n\}$ is increasing and $a_n\le 1$, for all $n\in\mathbb N$, and hence convergent to some positive $a\le 1$. (In fact, it converges to 1.)

Then $$ 2^{na_n}\le 2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}\le n\cdot 2^{na_n} $$ and hence $$ 2^{a_n}= \sqrt[n]{2^{na_n}}\le \sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}\le 2^{a_n}\sqrt[n]{n} $$ But $\,\,2^{a_n}\to 2^a$ and $\,\,\sqrt[n]{n}2^{a_n}\to 2^a$, and hence $$ \sqrt[n]{2^{na_n}}\le \sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}} \to 2^a. $$ Thus A is the correct answer.

Note. In fact $a=1$, and thus $$ \sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}\to 2. $$ This is due to Weyl's Equidistribution Criterion.

Answer 2

not a solution

Here is $n=2$ to $n=1000$.

Visual impression: converges to $2$, but not monotonically.

Answer 3

As @Jam mentioned in comments and GEdgar in his answer, it is not monotonic at all as shown in the table for the very first values of $n$ (put them on a scatter plot) $$\left( \begin{array}{cc} n & x_n \\ 1 & 1.79188 \\ 2 & 2.59580 \\ 3 & 2.39393 \\ 4 & 2.22146 \\ 5 & 2.12745 \\ 6 & 2.07215 \\ 7 & 2.07967 \\ 8 & 2.19713 \\ 9 & 2.16297 \\ 10 & 2.13347 \end{array} \right)$$ To extend GEdgar plot, consider $n=10^k$ and the values are $$\left( \begin{array}{cc} k & x_{10^k} \\ 1 & 2.133472777 \\ 2 & 2.030634143 \\ 3 & 2.005381446 \\ 4 & 2.000775098 \\ 5 & 2.000100408 \\ 6 & 2.000012379 \end{array} \right)$$

From these numbers, it seems that, more or less, $$\log(x_n-2)\sim 0.567936 -1.96092\, k \qquad (R^2=0.998712)$$