Limit of $\sum \limits_{k=1}^{n} \frac{2^kn+2n^2+k}{2^{k+1}n^2+2^kk}$ when $n\to\infty$

Question

I have to show the convergence of the series $$\lim\limits_{n \to \infty}a_n=\sum \limits_{k=1}^{n} \frac{2^kn+2n^2+k}{2^{k+1}n^2+2^kk}.$$ I am quite sure that the limit is 1.5. I wanted to show this by estimation for another series smaller than $a_n$ and a further one that is bigger than $\lim\limits_{n \to \infty}a_n$. I have been trying to find this estimation for hours without success. Can someone help me, please?

Answer 1

Fine as they are, the other answers gloss over an important nuance: In a comparison of (positive) sequences, if $a_n\le b_n$ and $\lim_{n\to\infty}b_n=L$, we cannot conclude that $\lim_{n\to\infty}a_n=L$. In fact, we cannot even conclude that $\lim_{n\to\infty}a_n$ exists (unless $L=0$). We need to squeeze $a_n$ between two sequences that both converge to the same limit.

To do this, we begin with

$$ {2^kn+2n^2+k\over2^{k+1}n^2+2^kk} ={2^kn+(2n^2+k)\over2^k(2n^2+k)} ={n\over2n^2+k}+{1\over2^k} $$

The key is to note that, for $1\le k\le n$,

$${1\over2n+1}={n\over2n^2+n}\le{n\over2n^2+k}\lt{n\over2n^2}={1\over2n}$$

and thus

$${1\over2n+1}+{1\over2^k}\le{2^kn+2n^2+k\over2^{k+1}n^2+2^kk}\lt{1\over2n}+{1\over2^k}$$

From this we see that

$${n\over2n+1}+\sum_{k=1}^n{1\over2^k}=\sum_{k=1}^n\left({1\over2n+1}+{1\over2^k}\right) \le a_n\lt\sum_{k=1}^n\left({1\over2n}+{1\over2^k}\right)={1\over2}+\sum_{k=1}^n{1\over2^k}$$

and now we see that $a_n$ is squeezed between two sequences that both tend to ${1\over2}+1=1.5$.

Answer 2

break it up.

$a_n=\sum \limits_{k=1}^{n} \frac{2^kn}{2^{k+1}n^2+2^kk} + \sum\limits_{k=1}^{n} \frac{2n^2}{2^{k+1}n^2+2^kk}+\sum\limits_{k=1}^{n} \frac{k}{2^{k+1}n^2+2^kk}.$

Since every element of each of those series is greater than zero, the series for $a_n$ converges iff all three of the above converge, and diverges if any of the above diverge.

$\sum \limits_{k=1}^{n} \frac{2^kn}{2^{k+1}n^2+2^kk} <\sum\limits_{k=1}^{n} \frac{1}{2n}$

$\sum\limits_{k=1}^{n} \frac{2n^2}{2^{k+1}n^2+2^kk} <\sum\limits_{k=1}^{n} \frac{1}{2^{k}}$

$\sum\limits_{k=1}^{n} \frac{k}{2^{k+1}n^2+2^kk} < \sum\limits_{k=1}^{n} \frac{1}{2^{k}}$

Answer 3

Factor $n^2$ out of the numerator and denominator to get $$a_n = \sum_{k=1}^n\frac{2 + \frac{2^k}{n} + \frac{k}{n^2}}{2^{k+1} + \frac{2^kk}{n^2}}.$$ Making the denominator smaller makes the whole fraction bigger, so you have the comparison $$a_n\leq \sum_{k=1}^n\frac{2+\frac{2^k}{n} + \frac{k}{n^2}}{2^{k+1}}.$$ Writing everything over a common denominator and performing some simple algebra, you get $$a_n\leq \sum_{k=1}^n\frac{1}{2^k} + \frac{1}{2} + \frac{1}{n^2}\sum_{k=1}^n\frac{k}{2^{k+1}}.$$ The first sum on the right hand side is a geometric series, which converges as $n\to\infty$, the second term is a constant, and for the third we can use that $k\leq 2^k$ when $k\geq 1$ to get the comparison $$a_n\leq \sum_{k=1}^n\frac{1}{2^k} + \frac{1}{2} + \frac{1}{2n}.$$ Taking the limit as $n\to\infty$ as using the algebraic limit laws, since each of the three terms is convergent, we get $$\lim_{n\to\infty}a_n = \sum_{k=1}^{\infty}\frac{1}{2^k} + \frac{1}{2} + \lim_{n\to\infty}\frac{1}{2n}$$ which will give you 1.5 as you claimed since the geometric series converges to 1 and $\frac{1}{2n}\to 0$ as $n\to\infty$.

EDIT: As pointed out in the comments, this doesn't actually show you what the limit is, rather bounds it above by 1.5, provided it exists.